
Class 

Book. F8 a 

Copyright N°_Jia_ 

CDEXRIGHT DEPOSfT. 



CAMS 

ELEMENTARY AND ADVANCED 



BY 

FRANKLIN DeRONDE FURMAN, M.E. 

Professor of Mechanism and Machine Design 

at Stevens Institute of Technology 

Member of American Society of Mechanical Engineers 



ELEMENTARY CAMS 
FIRST EDITION (THIRD IMPRESSION) 

TOTAL ISSUE FOUR THOUSAND 

CAMS-ELEMENTARY AND ADVANCED 
FIRST EDITION 



NEW YORK 

JOHN WILEY & SONS, Inc. 

London: CHAPMAN & HALL, Limited 
1921 



3®° 



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Copyright, 1916, 1921 

BY 

FRANKLIN DeRONDE FURMAN 



FEB -3 !32i 



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PRESS OF 

BRAUNWORTH & CO. 

BOOK MANUFACTURERS 

BROOKLYN. N. Y. 



©CU608208 



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PREFACE TO THE ENLARGED EDITION 

The first five sections of this book were published about three 
years ago under the title of "Elementary Cams." The chief features 
of this earlier book were that it pointed out a classification, an arrange- 
ment, and a general method of solution of the well-known cams in 
such manner as has been generally developed in other specialized 
branches in technical engineering work; and also it gave a series of 
cam factors for base curves in common use, which enabled designers 
to compute proper cam sizes for specific running conditions, offering 
numerous examples in the use of these factors in the several kinds of 
cam problems. The factors, with the exception of the one for the 
30° pressure angle for the Crank Curve were new, so far as the 
author is aware. The " Elementary Cams " will continue to be 
sold as a separate volume. 

A further development of the subject is given in the present work 
which is under the title of "Cams." The chief original features of 
this advanced work include the development or use, or both, of the 
logarithmic, cube, circular, tangential and involute base curves, 
the establishing of cam factors for such of these curves as have 
general factors, and the demonstration that the logarithmic base 
curve gives the smallest possible cam for given data. 

The new material now introduced into the book includes, further, 
comparisons of the characteristic results obtained from all base 
curves, in which the relative size of each cam, and the relative velocity 
and acceleration produced by each, is shown graphically in one 
combined group of illustrations, thus enabling the designer to glance 
over the entire field of theoretical cam design and quickly select the 
type that is best adapted for the work in hand. From these diagrams 
one may observe, for example, which form of cam is best adapted for 
gravity, spring or positive return, which is best for slow or fast veloci- 
ties at various points in the stroke, and which ones are apt to develop 
"hard spots" in running. The involute curve is found to have its 
chief and characteristic theoretical advantage when it is used with 
an offset follower. The nature of the contact between cylindrical, 
conical and hyperboloidal roller pins, when used in connection with 
grooved cylindrical cams, has been investigated and pointed out. 
The subject of pure rolling contact between various forms of oscillat- 
ing cam arm surfaces, and of the nature and amount of sliding action 

iii 



iv PREFACE TO THE ENLARGED EDITION 

of such surfaces has been developed so that the effects of wear due 
to rubbing may be confidently considered when such types of cams 
are under design. 

While the whole purpose of this work has been to present the 
subject matter in graphical form and in the simplest possible manner 
so as to make it available to the greatest number, much mathematical 
investigation has been necessary and in this I have been greatly 
aided by my colleague Professor L. A. Hazeltine, M. E., head of the 
department of Electrical Engineering at Stevens, to whom I express 
my deep appreciation. The details of these investigations are not 
necessary here and are not set down, but their results are. These 
results are given in various formulas that are used in the solution of a 
number of the problems. These final formulas avoid the use of 
calculus and are mostly in such form as to be readily used by designers 
generally. 

In closing, the author desires to introduce a personal thought 
that has grown up, and which is inseparable, with this book. Some 
years ago, before any special study was given by the writer to the sub- 
ject of cams, it appeared that the whole subject of mechanism was so 
thoroughly covered by various text books and technical papers that 
the time in engineering development had arrived when there was 
but little for an instructor to look forward to in the way of produc- 
tion of extended original work on any given topic. To say the least 
such a thought was not at all encouraging, and so it is a pleasure now 
to the author, and it is hoped that it will be an inspiration particularly 
to the younger readers, to record that the study of this subject of 
cams has brought forth a great wealth of new and practical material 
which had not previously been brought to light and set down in the 
literature of the subject. Now that this work is done, the vastness 
of the " unknown," even in this present era of great accomplish- 
ments, is realized as it never was before, and it only remains to suggest 
that not only this topic of cams but many other topics in the science 
of engineering may offer opportunities for much further development 
and perfection on the part of those who have the desire for such work 
and the time to pursue it. 

F. DeR. Furman. 
Hoboken, N. J., April, 1920. 



CONTENTS 



PAGES 

Section I. — Definitions and Classification 1-19 

Cams Follower Surfaces Radial or Disk Cams Side or 
Cylindrical Cams Conical and Spherical Cams 

Names of Cams — Periphery, Plate, Heart, Frog, Mushroom, Face, 
Wiper, Rolling, Yoke, Cylindrical, End, Double End, Box, Internal, 
Offset, Positive Drive, Single Acting, Double Acting, Step, Adjustable, 
Clamp, Strap, Dog, Carrier, Double Mounted, Multiple Mounted, 
Oscillating 

Definitions of Terms Used in the Solution of Cam Problems — Cam 
Chart, Cam Chart Diagram, Time Chart, Base Curve, Base Line, 
Pitch Line, Pitch Circle, Pitch Surface, Working Surface, Pitch Point, 
Pressure Angle 

Formula for Size of Cam for a Given Maximum Pressure Angle 

Table of Cam Factors for All Base Curves for Maximum Pressure 
Angles from 20° to 60° 

Section II. — Method of Construction of Base Curves in Common 

Use 20-24 

Straight Line Base Straight-Line Combination Curve Crank 
Curve Parabola Elliptical Curve 

Section III. — Cam Problems and Exercise Problems 25-74 

Problem 1, Empirical Design Problem 2, Technical Design. 
Advantages of Technical Design Problem 3, Single-Step Radial 
Cam, Pressure Angle Equal on Both Strokes Omission of Cam 
Chart Problem 4, Single-Step Radial Cam, Pressure Angles Unequal 
on Both Strokes 

Pressure Angle Increases as Pitch Size of Cam Decreases Change 
of Pressure Angles in Passing from Cam Chart to Cam Cam Con- 
sidered as Bent Chart. Base Line Angles Before and After Bending 

Limiting Size of Follower Roller Radius of Curvature of Non- 
Circular Arcs 

Problem 5, Double-Step Radial Cam Determination of Maximum 
Pressure Angle for a Multiple-Step Cam 

Problem 6, Cam with Offset Roller Follower Problem 7, Cam 
with Flat Surface Follower Limited Use of Cams with Flat Surface 
Followers 

Problem 8, Cam with Swinging Follower Arm, Roller Contact — 
Extremities of Swinging Arc on Radial Line Problem 9, Cam 
with Swinging Follower Arm, Roller Contact — Swinging Arc, Con- 
tinued, Passes Through Center of Cam Effect of Location of 
Swinging Follower Arm Relatively to the Cam 



VI CONTENTS 



PAGES 

Problem 10, Face Cam with Swinging Follower Problem 11, 
Cam with Swinging Follower Arm, Sliding Surface Contact Data 
Limited for Followers with Sliding Surface Contact 

Problem 12, Toe and Wiper Cam Modifications of the Toe 
and Wiper Cam 

Problem 13, Single Disk Yoke Cam Limited Application of 
Single Disk Yoke Cam Problem 14, Double Disk Yoke Cam 

Problem 15, Cylindrical Cam with Follower that Moves in a 
Straight Line Refinements in Cylindrical Cam Design Prob- 
lem 16, Cylindrical Cam with Swinging Follower Chart Method 
for Laying Out a Cylindrical Cam with a Swinging Follower Arm 

Exercise Problems, 3a to 16a 

Section IV. — Timing and Interference of Cams 75- 78 

Problem 17, Cam Timing and Interference Location of Key- 
ways Exercise Problem 17a 

Section V. — Cams for Reproducing Given Curves or Figures . . 79- 87 
Problem 18, Cam Mechanism for Drawing an Ellipse Prob- 
lem 18a, Exercise Problem for Drawing Figure 8 Problem 19, 
Cam for Reproducing Handwriting Using Script Letters Ste 

Method of Subdividing Circles into Any Desired Number of Equal 
Parts 

Section VI. — Advanced Group of Base Curves 88-137 

Complete List and Comparison of Base Curves, Their Appli- 
cations and Characteristic Motions Velocity and Acceleration 
Diagrams Showing Characteristic Action of Various Cams All- 
logarithmic Curve Gives Smallest Possible Cam for a Given Pres- 
sure Angle 

Problem 20, All-logarithmic Cam General Analysis Detail 
Construction of Logarithmic Curve and Cam by Analytical and 
Graphical Methods 

Problem 21, Logarithmic-combination Cam with Parabolic 
Easing-off Arcs 

Problem 22, Cam with Straight-Line Base 

Straight-Line Combination Curve Crank Curve Effect 
of Crank Curve Following Its Tangent Closely Parabola 
Gravity Curve Curve of Squares Perfect Cam Action 
Comparison of Parabola and Crank Curves 

Problem 23, Tangential Cam, Case 1 Graphical and Ana- 
lytical Methods Characteristic Retardation 

Problem 24, Circular Base Curve Cam, Case 1 Elliptical Base 
Curve Effect of Varying Axes of Ellipses Elliptical Base 
Curve Equivalent to Nearly All Other Base Curves 

Cube Curve Symmetrically and Unsymmetrically Applied 

Problem 25, Cube Curve Cam, Case 1 



CONTENTS Vll 



PAGES 



Cams Specially Designed for Low-Starting Velocities 
Problem 26, Circular Base Curve Cam, Case 2 
Problem 27, Cube Curve Cam, Case 2 
Problem 28, Tangential Cam, Case 2 

Section VII. — Cam Characteristics 138-156 

Methods of Determining Velocities and Accelerations Time- 
Distance, Time-Velocity and Time-Acceleration Diagrams 
Degree of Precision Obtained by Graphical Methods Comparison 
of Relative Velocities and Forces Produced by Cams of Different 
Base Curves Cam Follower Returned by Springs Rela- 
tive Strength of Spring Required for Cams of Different Base Curves 
Special Adaptation of Cube Curve Cam for Follower Returned 
by a Spring Na'ure of Pressure between Cam Surface and 
Spring-Returned Follower 

Accuracy in Cam Construction 

Regulation of Noise High-Speed Cams Balancing of Cams 
Pressure Angle Factors, Nature of Application and Method of 
Determination for All Base Curves Varied Forms of Funda- 
mental Base Curves Chart Showing Values of Intermediate 
Pressure Angles from 20° to 60° for all Cams 

Section VIII. — Miscellaneous Cam Actions and Constructions . 157-229 

Variable Angular Velocity in Driving Cam Shaft 

Problem 29, Oscillating Cam Having Variable Angular Velocity, 
Toe and Wiper Type 

Problem 30, Wiper Cam Operating Curved-Toe Follower 

Problem 31, Sliding Action between Cam and Flat Follower Sur- 
faces Rate of Sliding Measured Velocity of the Folbwer 
Measured 

Problem 32, Sliding Action between Cam and Curved-Toe Fol- 
lower 

Problem 33, Sliding Action where Driving Cam Ha^ Variable 
Angular Velocity 

Elimination of All Sliding Action between Cam and Flat or 
Curved Surface Follower 

The Princip e of Pure Rolling Action between Cam Surfaces 
Well- Known Curves that Lend Themselves Readily to Pure 
Rolling Cam Action 

Problem 31, Pure Rolling with Flat-Surface Follower 

Use of Logarithmic Curve for Pure Rolling Action Charac- 
teristic Properties of the Logarithmic Curve 

Problem 35, Pure Rolling with Logarithmic Curved Cam Arm 
Angular Motion of Each Arm Tangency of Logarithmic Cam 
Surfaces Regulation of Pressure Angle when Logarithmic 
Rolling Cams are Used 

Derived, or Computed Curves for Rolling Cam Arms 



Vlll CONTENTS 

PAGES 

Problem 36, The Use of a Derived Curve for Rolling Cam Arms 174 

Rolling Cam Arms Useful for Starting Shafts Gradually 

Regulation of Pressure Angle with Derived Rolling Cams 

Elliptical Arcs for Pure Rolling Cam Arms 

Problem 37, Elliptical Rolling Cam Arms, Angles of Action Equal 
Determination of Major and Minor Axes of Ellipses Construc- 
tion of Ellipse Pressure Angle in Rolling Elliptical Cam Arms 

Problem 38, Elliptical Rolling Cam Arms, Angles of Action 
Unequal 

Pure Rolling Parabolic Cam Surfaces for a Reciprocating Motion 

Problem 39, Rolling Parabolas Construction of Parabola 
Pure Rolling Hyperbolic Cam Arms where Centers are Close 
Together 

Problem 40, Rolling Hyperbolas Construction of Hyperbola 

Detail Drawing of Cylindrical Cams The True Maximum 
Pressure Angle in Cylindrical Cams Drawing of Groove Out- 
lines, Approximate and More Exact Methods 

Forms of Follower Pins for Cylindrical Grooved Cams Line of 
Contact between Pin and Groove Surface, at Rest and Moving 
The Cylindrical Follower Pin The Conical Follower Pin The 
Hyperboloidal Follower Pin 

Plates for Cylindrical Cams Adjustable Cylindrical Cams for 
Automatic Work 

Double-Screw Cylindrical Cams Periods of Rest of More 
than One Revolution in Cylindrical Cams Slow-advance and 
Quick-Return Secured by Double-Screw Cam 

Straight-Sliding Plate Cams 

Involute Cams Construction of Involute Curve Pressure 
Angle with Involute Cam 

Involute Cam Specially Adapted for Flat-Surface Follower 

Problem 41, Involute Cam with Radial Follower 

Oscillating Positive-Drive-Single-Disk Cam Cam Shaft Acting 
as Guide Positive Drive with Cam Shaft as Guide Positive- 
Drive Double-Disk Radial Cam with Swinging Follower Rotary- 
Sliding Yoke Cams Giving Intermittent Harmonic Motion, and 
Reciprocating Motion Rotary Sliding Yoke Cam, General Case 
Cam Surface on Reciprocating Follower Rod 

Problem 42, Definite Motion where Cam Surface is on Follower 
Rod 

Problem 43, Cam Surface on Swinging Follower Arm 

Effect of Swinging Transmitter Arm between Ordinary Radial 
Cam and Follower Angular Velocity Curve for a Swinging 
Follower Arm Velocity Curve for a Follower Rod with Com- 
parison of Results Obtained by Using Transmitter Arms with 
Sliding and Roller Action Diagram of Pressure Angles Meas- 
urement of Rubbing Velocities in Cams Having Sliding Action 
Boundary of Follower Surface Subjected to Wear in Sliding Cams 221 



CONTENTS IX 

PAjGES 

Cam Action Different on Forward and Return Strokes with Sliding 222 

Cams 

Problem 44, Small Cams with Small Pressure Angles Secured by 
Using Variable Drive Variable Drive by Whit worth Motion 

Swash Plate Cam Uniformly Rotating Cam Giving Inter- 
mittent Rotary Motion The Eccentric a Special Type of Cam 
An Example of a Time-Chart Diagram for Eleven Cams on One 
Shaft of an Automatic Machine 229 



; 



ELEMENTARY GAMS 



SECTION I.— DEFINITIONS AND CLASSIFICATION 

Definitions 

1. Cams are rotating or oscillating pieces of mechanism having 
specially formed surfaces against which a follower slides or rolls 
and thus receives a reciprocating or intermittent motion such as 
cannot be generally obtained by gear wheels or link motions. 

Various forms of cams are illustrated at C in Figs. 1 to 10. The 
follower in each case is shown at F, all having roller contact except 
the ones shown in Figs. 7 and 8. The former has a V edge and 
the latter a plane surface in contact with the cam and both have 
sliding action. 

2. Follower edges or rollers may have motion in a straight 
line as from D to G, Fig. 7, or in a curved path depending on suit- 
ably constructed guides or on swinging arms. The total range of 
travel of the follower may be accomplished by one continuous 
motion, or by several separate motions with intervals of rest. Each 
motion may be either constant or variable in velocity, and the time 
used by the motion may be greater or less, all according to the 
work the machine has to do and to the will of the designer. 

Classification 

3. Cams may be most simply, and at the same time most com- 
pletely, classified according to the motion of the follower with re- 
spect to the axis of the cam, as: 

(a) Radial or disk cams, in which the radial distance from the 
cam axis to the acting surface varies constantly during part or all of 
the cam cycle, according to the data. The follower edge or roller 
moves in all cases in a radial, or an approximately radial, direction 
with respect to the cam. Various forms of radial cams are illus- 
trated in Figs. 1, 2, 7, 8, and 9. 

(b) Side or cylindrical cams, in which the follower edge or 
roller moves parallel to the axis of the cam, or approximately in 

1 



ELEMENTAEY CAMS 



/ 




F 






s " 








this direction. Several types of side cams are shown in Figs. 3, 4, 
and 10. 

Nearly all the cams referred to in the above figures illustrating 
the two general classes of radial and side cams respectively have 
special or local trade names which will be pointed out in a succeed- 
ing paragraph. 

(c) Conical and (d) Spherical cams, in which the follower edge 
or roller moves in an inclined direction having both radial and 

longitudinal components 
with respect to the axis 
of the cam as illustrated 
in Figs. 5 and 6. 

4. Names of cams. 
Cams, in popular usage, 
have come to be known 
by a wide range of names, 
the same cam often being 
designated by a number 
of different names accord- 
ing to geographical loca- 
tion and personal prefer- 
ence and surroundings of 
the cam builder or user. 
This is an unfortunate con- 
dition, and in the general 
classification in the preced- 
ing paragraph an endeavor 
is made to establish a fun- 
damental basis for clarifying and simplifying the nomenclature 
of cams as much as possible. In a treatise of this kind, however, it 
is essential that, at least, the more common of the ordinary working 
terms be recognized and defined, and that the cams under their 
popular names be properly placed in the fundamental classification 
given in the preceding paragraph. 

The following specially named cams fall under the classifica- 
tion of radial cams: 

(e) Periphery cams, in which the acting surface is the periphery 
of the cam, as illustrated in Figs. 1, 7, and 9. While these are ex- 
amples of true periphery cams, it must be recorded that the cylin- 
drical grooved cam, shown in Fig. 3, is also known to some extent 
as a periphery cam, due no doubt to the fact that in designing this 





' 









r 


1 


\ 




— 


1 


.._. 
































i 






i 






/ 










c 





End 



Front 



Fig. 1. 



-Radial Cam and Follower, 
Roller Contact 



DEFINITIONS AND CLASSIFICATION 3 

cam the original layout for the contour of the groove is first made 
on a flat piece of paper, which is then wrapped on to the surface or 
"periphery" of the cylinder. Since the contour line of the groove 
which lies on the periphery is merely a guiding line for cutting the 
groove, and since the side surface of the groove is the working sur- 
face, it is, to say the least, a misnomer to designate such a cam as 
a periphery cam. 

(/) Plate cams, in which the working surface includes the full 
360°, and forms either the periphery of the cam, or the sides of a 




E>'D FlJOXT 

Fig. 2. — Face Cam and Follower 



groove cut into the face of the cam plate, as illustrated in Figs. 1 
and 2 respectively. Figs. 7 and 9 also show plate cams. 

(g) Heart cams, in which the general form is that which the 
name implies. See Fig. 7. In this type of cam there are two 
distinct symmetrical lobes, often so laid out as to give uniform 
velocity to the driver. In this case each lobe would be bounded 
by an Archimedean spiral with the ends eased off. 

(h) Frog cam, in which the general form includes several lobes 
more or less irregular, as illustrated, for example at C in Fig. 9. 

(i) Mushroom cam, in which the periphery of a radial or disk 
cam works against a flat surface, usually a circular disk at right 
angles to the cam disk, instead of against a roller, see Fig. 44. 

0') Face cam, also called a Groove, but more properly a Plate 
Groove cam, to distinguish it from the Cylindrical Groove cam, in 
which a groove is cut into the flat face of the cam disk. In 



4 ELEMENTARY CAMS 

this form of cam shown in Fig. 2 the roller has two opposite lines 
of contact, one against each side of the groove, when the roller has 
a snug fit. The plate or disk in which the groove is cut is generally 
circular; but it may be cast to conform with the contour of the 
groove, or it may be built with radial arms supporting the irregular 
grooved rim. In the latter case it lacks resemblance to the face 




Top 




End 



Front 
Fig. 3. — Cylindrical Cam and Swinging: Follower 



cam, but nevertheless it must, because of the nature of its action, 
be classed with it. The face cam, as ordinarily considered and as 
illustrated in Fig. 2, is better adapted for higher speeds because of 
its more nearly balanced form of construction. Against this, how- 
ever, must be considered one of two disadvantages, either the high 
rubbing velocity of the roller against one side of the groove when 
the roller is a snug fit, or lost motion and noise as the working line 
of contact changes from one side of the groove to the other when 
the roller has a loose fit. The most important advantage of the 
face cam, that of giving positive drive, will be considered in para- 
graph 9. The term groove cam might be applied, with advantage 
in clearness of meaning, to such face cams as are cut or cast on 
non-circular plates. 



DEFINITIONS AND CLASSIFICATION 



(k) Wiper cam, which has an oscillating motion, and is con- 
structed usually with a long; curved arm in order that it may "wipe" 
or rub along- the plane surface of a long projecting "toe," or follower. 
The wiper cam is used generally to give motion to a follower which 
moves straight up and down as shown from F to F' in Fig. 8. This, 
however, is not essential and the follower may also have a swinging 




Fig. 4. — End Cam and Follower 

motion. The disadvantage of sliding friction, which is inseparable 
from the wiper cam, is balanced to some extent by the fact that 
the very sliding permits, within certain range, of the assignment of 
specified intermediate velocities between the starting and stopping 
points which cannot be obtained with similar forms of cams which 
have pure rolling action. 

(/) Rolling cam, which greatly resembles the wiper cam in 
general appearance, but which is totally different in principle, for 
the curves of the cam and follower surfaces are specially formed so 
as to give pure rolling action between them. The rolling cam is 
specially well adapted to cases where both driver and follower have 
an oscillating motion and where the velocities between the starting 
and stopping points are not important and are not specified. 



6 



ELEMENTARY CAMS 



(m) Yoke cam, a form of radial cam in which all diametral lines 
drawn across the face and through the center of rotation of the 
cam are equal in length. This form of cam permits the use of 
two opposite follower rollers whose centers remain a fixed distance 
apart, to roll simultaneously on opposite sides of the cam, and thus 
give positive motion to the follower. For illustration, see Fig. 9. 





Fig. 5. — Conical Cam and Recipro- 
cating Follower 



Fig. 6. 



-Spherical Cam and Swinging 
Follower 



Yoke cams may be, and frequently are, made of two disks fixed 
side by side, the peripheries being complementary to each other 
and the two rollers of the yoke rolling on their respective cam surfaces, 
as shown in Fig. 56. The advantage of yoke cams is that they 
give positive motion with pure rolling of the follower roller, there 
being contact on only one side of the roller in contradistinction to 
the double contact of th^ roller which exists in face and groove 
cams. 

5. The following specially named cams fall under the general 
classification of side cams. 

These include cams that have been made from blank cylindrical 
bodies by using a rotary end cutter with its axis at right angles 
to the axis of the cylinder and by moving the axis of the rotary 
cutter parallel to the axis of the cylinder while the cylinder rotates. 
A groove of desired depth is thus left in the cylinder, Fig. 3, or the 
end of a cylindrical shell is thus milled to a desired form, Fig. 4. 
A side cam may also be formed by screwing a number of formed 



DEFINITIONS AND CLASSIFICATION 



clamps on to a blank cyl- 
inder, the sides of the 
clamps thus acting as the 
working surface as illus- 
trated in Fig. 11. All 
types of side cams may 
properly be considered 
as derived from blank 
cylindrical forms, and, 
therefore, the name "cyl- 
indrical cam" could be 
regarded as synonymous 
with side cam; but gen- 
eral custom has limited 
the use of the term cyl- 
indrical cam to the "bar-" 
rel" or "drum" type 
mentioned below: 

(n) Cylindrical cam, 
also called Barrel cam, 
Drum cam, or Cylindrical 
Groove cam, in which the fig. 7 
groove, cut around the 

cylinder, affords bearing surface to the two opposite sides of the 

follower roller, thus giving positive motion, as illustrated in Fig. 3. 

(o) End cam, in which the working surface has been cut at the 

end of a cylindrical shell, thus re- 
quiring outside effort such as a 
spring or weight to hold the follower 
roller against the cam surface during 
the return of the follower. An end 
cam is shown in Fig. 4. 

(p) Double end cam, in which 
a projecting twisted thread has been 
left on a cylindrical body, against 
both sides of which separate rollers 
on a follower arm may operate, 
and thus secure positive motion. 
Instead of cutting down a cylinder 
to leave a projecting twisted thread, 
fig. s.— toe and wiper cam it may be cast integral with a 




-Heart Cam and Follower, Sliding Contact 




8 



ELEMENTARY CAMS 



warped plate, as illustrated in Fig. 10, but this in no way changes 
its characteristic action. 

There are a number of names in common use for cams, that 
cover both radial and side cams. Most prominent in this connection 
are those mentioned in paragraphs 6 to 14. 

6. Box cam, which designates a cam in which the follower roller 
is encased between two walls as in the face cam, Fig. 2, or the cylin- 
drical cam, Fig. 3. Literally, box cams would also include yoke 
cams, in which the yoke would be the "box." Box cams, because of 
their form of construction, give a positive drive in all cases. 

7. Internal cam, in which there is only one working surface, 
and this is outside of the pitch surface. The internal cam cor- 
responds to the internal gear wheel in toothed gearing. It may also 
be considered as a face cam with the inside surface of the groove 
removed, thus requiring that the follower roller should always be in 
pressure contact on the outside surface of the groove by means of a 
spring or weight, etc. Under some conditions of structural arrange- 
ments of the cam machine, the internal cam may be used to advan- 
tage where it will give a positive motion to a follower on the opposite 
stroke to that of the periphery cam; and it will also sometimes 




Fig. 9. — Yoke Cam 



permit of a larger roller than the periphery cam, as explained in 
paragraphs 56 and 62. 

8. Offset cam, in which the line of action of the follower, 
when extended, does not pass through the center of the cam, 
see Fig. 43. 

9. Positive-drive cam is one in which the cam itself drives 
the follower on the return as well as the forward motion. Most 



DEFINITIONS AND CLASSIFICATION 



9 



cams drive only on the forward motion of the follower and depend 
upon gravity or the action of a spring to drive the follower in its 
return motion; such cams are illustrated in Figs. 1, 4, 5, 6, 7, and 8. 
Cams having positive drive, and therefore independent of gravity 
or springs, are illustrated in Figs. 2, 3, 9, and 10. It will be noted 
that positive-drive cams include the face, yoke, cylindrical, and 
double-end types of cams; also that the box cam, although it in- 
cludes some of these, should also be considered as a group name of 
the positive-drive type. 

10. Single-acting and double-acting cams comprise all forms 
of cams, the single-acting ones giving motion only in one direction 
and depending on a spring or gravity to return the follower. Double- 
acting cams have the follower under direct control all the time and 
are the same as positive-drive cams described in the preceding 
paragraph. 

11. Step cams. Cams which give continuous motion to the 



llkiTH^ 




Fhont 



Fig. 10. — Double-End Cam 



follower from one end of the stroke to the other are called single- 
step cams. When the follower's motion in either of its two general 
directions is made up of two entirely separate movements it is called 
a double-step cam with reference to that stroke. If three or more 
separate movements are given to the follower while it moves in one 
general direction it is generally referred to as a multiple step cam, 
or as a triple-step, quadruple-step cam, etc. Since a cam may be, 
for example, a double-step cam on the out or working stroke, and 




Front 



,End 



Fig. 11. — Barrel Cam 




Fron,t End 

Fig. 12. — Adjustable Plate Cam 




Front 

Fig. 13. — Dog Cam 



DEFINITIONS AND CLASSIFICATION 11 

a single-step cam on the return stroke, such a cam may be referred 
to as a two-one step cam, always giving the number referring to the 
working stroke first. 

12. Adjustable cam, also called clamp cam, strap cam, dog 
cam, and carrier cam, in which specially formed pieces are directly 
bolted or clamped to any of the regular geometrical surfaces, usually 
to either the plane or cylindrical surfaces. In Fig. 12 the clamps 
are shown at C and D fastened to a disk. The cam, considered as a 
whole, belongs to the radial class. In Fig. 13 the clamps are shown 
at C and D, also fastened to a disk, but in this case the clamps, or 
dogs, as they are usually called when used in this way, are so formed 
as to give a sidewise motion to the follower, and therefore this cam 
belongs to the side cam class. In Fig. 11 clamps are shown at C, 
D, E, and F fastened to a cylinder, and they are shaped to give the 
same action as a regularly formed end-cam in the side-cam class. 
The type of cam illustrated in Fig. 11 is also known as an adjustable 
cylindrical or "barrel" or "drum" cam and is very widely used for 
regulating the feeding of the stock, and in operating the turret in 
automatic machines for the manufacture of screws, bolts, ferrules, 
and small pieces generally that are made up in quantities. 

13. Double-mounted or multiple-mounted cams are some- 
times resorted to where several movements can be concentrated 
into small space. This consists merely in placing two or more of 
any of the cam surfaces described in the preceding paragraphs on 
one solid casting or cam body. For example, a face cam, a cylin- 
drical, and an end cam may all be cut on one piece. 

14. Oscillating cams, in which the cam itself turns through a 
fraction of a turn instead of through the entire 360°. While any 
type of cam may be designed to oscillate instead of rotate, it is 
usually the toe-and-wiper and rolling forms of the radial type of 
cam that are known as oscillating cams. With oscillating cams the 
follower may move forth and back on a straight line, or it may 
oscillate also. 

15. Cams falling in the conical class have no special name other 
than the one here used. The spherical cams are sometimes termed 
globe cams. Cams in conical and spherical classes are particularly 
useful in changing direction of motion in close quarters and in 
directions other than at right angles. In both Figs. 5 and 6, end 
action of the cam is shown, but it is apparent that with thicker walls 
on both the cone and the sphere, grooves could be cut in them, 
thus giving positive driving cams in both cases. 



12 



ELEMENTARY CAMS 



16. Summing up the general and special names for cams we 
have in tabular form: 

p. Periphery 
/ Plate 
g Heart 
h Frog 
i Mushroom 
j Face or Plate Grooved 
k Toe and Wiper 
I Rolling 

s m Yoke or Duplex 
n Cylindrical, Grooved, Barrel, or 

Drum 
o End 
p Double End 



Cams 



Box 

Internal 
Offset 

Positive Drive 
Single Acting 
Double Acting 
Step 
Adjustable or 

Strap 
Dog or Carrier 
Multiple 

Mounted 
. Oscillating 



a Radial 
or Disk 



Side, or. 
Cylindrical 



c Conical 

d Spherical 
or Globe 



Definitions of Terms Used in the Solution of Cam Problems 

17. Cam chart. Illustrated in Fig. 14. The chart is a rectangle 
the height of which is equal to the total motion of the follower in 
one direction, and the length equal to the circumference of the pitch 
circle of the cam. The chart length represents 360° and is sub- 



o 


J> H 






* 




X 




% 


1 


c 
S 




4 










^Se 


\ 


Pitch Line 






cLUMT 












?• 


V 10 20 30 40 50'60 70 '80^90° 




180" 




270" 


360- 


~ 






F 

















Fig. 14. — Cam Chart 



divided into equal parts marking the 5°, 10 c 



points, or the }/§, 



34 • • • points, or any other convenient subdivision, according to 
the requirements of the problem. On the cam chart are drawn the 
base curve and the pitch line. The former becomes the pitch surface 
of the cam and the latter the pitch circle. 

18. Cam chart diagram. Illustrated in Fig. 15. The cam 
chart diagram is a rectangle, the height of which represents the 
total motion of the follower in one direction. The length of the 
diagram represents the circumference of the pitch circle of the cam. 



DEFINITIONS AND CLASSIFICATION 



13 



In the cam chart diagram the scales for drawing the height and the 
length of the rectangle are totally independent of each other and 
independent also of the scale of the cam drawing. In drawing the 
diagram no scale need be used at all, and the entire chart diagram 
with its base curve and pitch line may be drawn entirely freehand 
with suitable subdivisions marked off entirely "by eye" according 
to the requirements of the problem. The base curve may be drawn 
roughly as a curve or it may be made up of a series of straight lines. 
The cam chart diagram frequently serves all the purposes of the 
cam chart. It saves time, and permits of chart drawings being 




It0° 270° 3 60 

■Represents tenft/i of circumference of pitch circ/e ofcam^ 

Fig. 15. — Cam Chart Diagram 



made on small available sheets of paper, w T hereas the more precise 
cam chart often requires large sheets of paper which are usually 
impracticable and unnecessary in many circumstances. 

19. Time charts. Illustrated in Figs. 16 and 17. Time charts 
are the same as cam charts or cam chart diagrams, and are con- 
structed in the same way as described in the two preceding para- 
graphs. The term "time chart," however, is most appropriately 
applied to problems where two or more cams are used in the same 
machine and where their functions are dependent on each other. 



,1 
















f 


fi 




N^ 








"""* 




^- 




____ 


















j'lT 


4 

T 

Yvy 










--!> 










c 


£' 


c 




U. 



0° 90° 180° 270° 36(r 

Fig. 10. — Time Chart Diagram, Base Curves Superposed 

The time chart permits of allowances being made for avoiding 
possible interference of the several moving parts, and for the desired 
timing of relative motions for each part. The time chart contains 
two or more base curves according to the number of cams used. 
When the base curves are superposed as in Fig. 16, the time chart 
consists of a single rectangle whose height is equal to the greatest 



14 



ELEMENTAKY CAMS 



follower motion. The superposing of curves and lines often leads 
to confusion and error, and it is better, in general, that the time 
chart should consist of a series of charts or rectangles all of the 
same length and one directly under the other as in Fig. 17. Where 
there are many base curves it is desirable to separate the rectangles 









\ 


1 $n v 




1 1 i*"*^ 


S^\ 



0° 90° 180° 270° 360 

Fig. 17. — Time Chart Diagram, Base Curves Separated 



by a small space to avoid any possibility of confusion due to different 
base curves running together. In many cases the term "time chart 
diagram," or " timing diagram," will be more appropriate than "time 
chart" in just the same way as the cam chart diagram is more ap- 
propriate than the cam chart. 

20. Base curve. Illustrated in Fig. 14. A base curve is made 
up of a series of smooth continuous curves, or a combination of 
curves and straight lines, which represent the motion of the follower, 
and which run in a wave-like form across the entire length of the 
cam chart or diagram. The base curve of the cam chart becomes 
the pitch surface of the cam. 

21. Base line. Illustrated in Fig. 15. A base line is made up 
of a series of inclined straight lines, or a series of inclined and hori- 
zontal lines, in consecutive order, which zigzag across the entire 
length of the chart. The base line when used on the cam chart 
indicates the exact motion of the follower, but when used on a cam 
chart diagram it is merely a time-saving substitute for the drawing 
of the base curve. The base line of the cam chart diagram represents 
the pitch surface of the cam. 

22. Names of base curves or base lines in common use, 
see Figs. 18 and 19: 

1. Straight line 4. Parabola. 

2. Straight-line combination 5. Elliptical curve. 

3. Crank curve. 



DEFINITIONS AND CLASSIFICATION 



15 



23. Pitch line. Illustrated in Fig. 14. A pitch line is a 
horizontal line drawn on the cam chart or diagram, and it becomes 
the pitch circle of the cam. The position, or elevation, of the pitch 
line on the chart varies according to the base curve which is specified, 
and according to the data of the problem. For cams which give a 




Fig. 18. — Comparison of Base Curves in Common Use Showing Varying Degrees 
of Maximum Slope When Drawn in Same Chart Length 

continuous motion to the follower during its entire stroke, or throw, 
the pitch line will pass through the point on the base curve which 
has the greatest slope, starting from the bottom of the chart. This 
does not apply to all possible base curves, but it does apply to all 




1 Straight Line 

2 Straight Line Combination 

3 Crank Curve 

4 Parabola 

5 Elliptical Curve 



Fig. 19. — Comparison of Base Curves in Common Use Showing Uniform Maximum 
Slope of 30° When Drawn in Charts of Varying Length 



those mentioned in the preceding paragraph, a minor exception 
being made of the crank curve which will be referred to in para- 
graph 34. When the cam causes the follower to move through its 
total stroke in two or more separate steps the position of the pitch 
line on the chart must be specially found as will be explained in 
problem 5. 



16 ELEMENTARY CAMS 

24. Pitch circle. Illustrated in Fig. 20. A pitch circle is drawn 
with the center of rotation of the cam as a center, and its circumfer- 
ence is equal to the cam chart length. Its characteristic is that it 
passes through that point A, Fig. 20, of the pitch surface of the cam 
where the cam has its greatest side pressure against the follower. 
This applies to all cams in which the center of the follower roller 
moves in a straight radial line. For other motions of the follower 
roller, and for flat-faced followers, the pitch circle must be specially 
considered, as will be explained in some of the problems covering 
these types. 

25. Pitch surface. Illustrated in Fig. 20. The pitch surface 
of a cam is the theoretical boundary of. the cam that is first laid 
down in constructing the cam. When the follower has a V-shaped 
edge, as at D in Fig. 7, the pitch surface coincides with the working 
surface of the cam. When the follower has roller contact, as in 
Fig. 20, the pitch surface passes through the axis of the roller and 
the working or actual surface of the cam is parallel to the pitch 
surface and a distance from it equal to the radius of the roller. 

26. Working surface. Illustrated in Fig. 20. The working 
surface of the cam is the surface with which the follower is in actual 
contact. It limits the working size and weight of cam. For exact 
compliance with a given set of cam data, the cam has only one 
theoretical size which is bounded by the pitch surface, but the 
working size may be anything within wide limits which depend on 
the radius of the follower roller and the necessary diameter of the 
cam shaft. 

The working surface is found by taking a compass set to the 
radius of the roller and striking a series of arcs whose centers are 
on the pitch surface. Such a series of arcs is shown in Fig. 20 with 
their centers at B, A, etc. The curve which is an envelope to 
these arcs is the working surface. 

27. Pitch point of follower. Illustrated in Fig. 20. The 
pitch point of the follower is that point fixed on the follower rod or 
arm which is always in theoretical contact with the pitch surface 
of the cam. If the follower has a sharp V-edge the pitch point is 
the edge itself. If the follower has a roller end, the pitch point 
is the axis of the roller. The pitch point is constantly changing its 
position from C to D as the follower moves up and down. 

28. Pressure angle. Illustrated in Fig. 20. The pressure 
angle is the angle whose vertex is at the pitch point of the follower 
in its successive positions and whose sides are the direction 



DEFINITIONS AND CLASSIFICATION 



17 



of motion of the pitch point and the normal to the pitch 
surface. 

Pressure angles exist when the surface of the cam presses sidewise 
against the follower; they cause bending in the follower arm and 
side pressure in the follower guide and in the bearings. The pres- 



Maximum Pressure 
Angle^-\ Normalto 

.Pitch Surface 



Radial 
Line 




Fig. 20. — Showing Names of Surfaces, Lines, and Points of a Cam 



sure angle is constantly varying in all cams as the follower moves 
up and down, except where a logarithmic spiral is used. In assign- 
ing cam problems the maximum permissible pressure angle is usually 
given. In Fig. 20 the pressure angle is zero at C, it will be equal to 
a when B reaches J, and will be a maximum when A reaches K. 

29. Formula for size of cam for a given maximum pressure 
angle. The radius of the pitch circle of the cam may be found 
directly by the formula: 

360 . . . . . 1 



hX- V XfX 2 



= 57.3 



hf 



or, 



^ X T X ' X A 



= .159 



hf 



(1) 



(2) 



18 



ELEMENTARY CAMS 



in which, r = radius of pitch circle of cam. 
h = distance traveled by follower. 
/ = factor for a given maximum pressure angle. 
b = angle, in degrees, turned by cam while follower moves 

distance h. 
e = angle, in fraction of revolution, turned by cam while 

follower moves distance h. 

30. Cam factors for maximum pressure angle. The factors, 
or value of /, for various maximum pressure angles for cams using 
the several base curves in common use are: 



Table of Cam Factors 





Maximum Pressure Angle 


and Values 


OF / 


Name of Base Curve 


20° 


30° 


40° 


50° 


60° 


Straight line 


2.75 
3.10 
4.32 
5.50 
6.25 


1.73 

2.27 
2.72 
3.46 
3.95 


1.19 

1.92 

1.87 
2.38 

2.75 


.84 
1.77 
1.32 
1.68 
1.95 


.58 


Straight-line combination* . . . 
Crank curve 


1.73 
.91 


Parabola 


1.15 


Elliptical curvef 


1.35 







These factors, for 30°, are illustrated in Fig. 19 where each of 
the base curves is given such a length, in terms of the height, that 
they will all have the "same maximum slope. The values given in 
this table are also shown, graphically, in Fig. 21, thus enabling one 
to find the proper cam factor for any intermediate pressure angle 
between 20° and 60°. 



* For case where easing off radius equals follower's motion. 

t For case where ratio of horizontal to vertical axes of ellipse is 7 to 4. 



DEFINITIONS AND CLASSIFICATION 



19 



60 



B L DQ F 



50 



10 



30 



20 







\ 


\ 


TVT 












































\ 


v\\ 














































\ 


\A\ 














































\ 






1, 












































\ 






A- 












































\ 


















































\ 
















































\ 




JV\ 














































V S 




\ 




A 






































tA 




s 




\*v 




































hr 








X" c 


k 






































<* 










u 


k%> 


































$ 




\<3, 






^ 


































\< 


&_P§ 










/c 


*, 
































^■•^V-4 






\ 




s < 


































'#' 










N 




































fV% 












^s 


































A. 
















































r? 












































Ss 






























"*->-* 



/■: 



M 



1 2 .4 3JS 

Cam Factors 
Fig. 21. — Chart Showing Relation Between Pressure Angles a.nd Cam Factors 
for the Ordinary Base Curves 



SECTION II.— METHOD OF CONSTRUCTION OF BASE 
CURVES IN COMMON USE 

31. Detail construction of base curves. The method of 
constructing the several base curves for a rise of one unit of the 
follower will be explained in the succeeding paragraphs. The curves 
will be constructed to give a pressure angle of 30° by selecting factors 
from the 30° column in the table in the preceding paragraph. Should 
the base curve for any other pressure angle be desired the factor 
should be taken from the corresponding column. 

32. Straight-line base. Fig. 22. Lay off A B equal to the 
follower motion, which will be taken as 1 unit in these illustra- 
tions. Multiply this by the factor 1.73 from paragraph 30, and 
lay off the distance A R equal to it. Complete the parallelogram 
and draw the diagonal. This will be the straight line base and the 




Fig. 22. — Straight Base Line 



Fig. 23. — Straight-Line Combination Curve 



angle R AC will be 30°. A R will be the pitch line. These base 
lines and curves are laid off from right to left so that they may be 
used in a natural manner later on in laying out the cam so that 
it will turn in a right-handed or clockwise direction. 

The straight-line base gives abrupt starting and stopping velocities 
at the beginning and end of the stroke and causes actual shock in 
the follower arm. The velocity of the follower during the stroke is 
constant. The acceleration at starting and retardation at stopping is 
infinite and is zero during the stroke. 

33. Straight-line combination curve. Fig. 23. Construct 
the rectangle with a height of 1 unit and a length of 2.27 units. 
With B and R as centers draw the arcs A E and C N, and draw a 
straight line E N tangent to them. The angle FEN will then equal 
30° and the line A C will be a base curve made up of arcs and a 

20 



CONSTRUCTION OF BASE CURVES IN COMMON I SE 



21 



straight line combined to form a smooth curve. DF will be the 
pitch line. 

The straight-line combination curve, being rounded off at the 
ends, does not give actual shock to the follower at starting and stop- 
ping, but it does give a more sudden action than any of the base 
curves which follow, and the maximum acceleration and retardation 
values are comparatively larger. 

34. Crank Curve. Fig. 24. Construct the rectangle. Draw 
the semicircle R G C and divide it into any number of equal parts. 
Six parts are best for practice work for this curve, but in general 
in practical work the greater the number of divisions the more 
accurate will be the curve and the smoother the action of the cam. 



5-^< 


^C 










Bk 


u r 






^ 












gI 


F 






^\U3 


I 


H 




D \ 


















gj£ 










K 




a\ 










k— 






















' 





Fig. 24. — Crank Curv; 



The six equal divisions of the semicircle are readily obtained by 
taking G as a center and F C as a radius and striking arcs at 1 and 
5, then with R and C as centers mark the points 2 and 4 respectively. 
Divide the length of the chart into six equal parts, as at H, I, E, etc. 
From these points drop vertical lines, and from the corresponding 
divisions on the semicircle draw horizontal lines, giving intersecting 
points, as at K, on the desired crank curve. The tangent to the 
curve at E will then make an angle of 30° with the line E F. The 
pitch line will be D F. 

When the crank curve is transferred from the chart to the cam 
it gives an angle which is a fraction of a degree greater than 30° 
at the point E on the cam in practical cases. This is not enough 
greater to warrant the special computations and drawing that would 
be necessary to be exact. Therefore the method of laying out the 
crank curve and the pitch line, as given above, will be adhered to 
in this elementary consideration of cam work, because of its 
simplicity. 

The crank curve gives a slightly irregular increasing velocity 
to the follower from the beginning to the middle of its stroke; then 
a decreasing velocity in reverse order to the end of the stroke. The 



22 



ELEMENTARY CAMS 



acceleration diminishes to zero at the middle of the stroke and then 
increases to the end. The maximum acceleration and retardation 
values are much less than for the straight-line combination curve, 
and are only a little greater than for the parabola. 

35. Parabola. Fig. 25. Construct the rectangle. Draw the 
straight line R S in any direction and lay off on it sixteen equal 
divisions to any scale. From the sixteenth division draw a line to F, 
the middle point of the chart; draw other lines parallel to this 
through the points 9, 4, and 1, thus dividing the distance R F into 
four unequal parts which are to each other, in order, as 1, 3, 5, and 
7. From these division points draw horizontal lines, and from H, 
I, and J drop vertical lines. The intersecting points, as at K, 




Fig. 25. — Parabola 



will be on the desired parabola. The points H, I, and J divide the 
distance D E into four equal parts. 

The parabola gives a uniformly increasing velocity from the 
beginning to the middle of the stroke; then a uniformly decreasing 
velocity to the end. The acceleration of the follower is constant 
during the first half of the stroke and the retardation is constant 
during the last half. The acceleration and retardation values are 
equal and are less than the maximum value of any of the other base 
curves. This means that the direct effort required to turn a positive- 
acting parabola cam is less than for any other type of positive cam. 

36. To better understand the smooth action given by the cam 
using this curve, consider, 1st, D H as a time unit during which the 
follower rises one space unit ; 2d, H I as an equal time unit during 
which the follower rises three space units; 3d, 7 J as the time unit 
during which the follower rises five space units, etc. Inasmuch as 
the follower travels two units further in each succeeding time unit, 
it gains a velocity of two units in each time unit, and this is uniform 
acceleration. 

The distance from F to C would be divided the same as from 
F to R and points on the part of the curve from E to C similarly 



CONSTRUCTION OF BASE CURVES IN COMMON USE 



23 



located. This curve will be identical with E A, but in reverse order, 
and will give uniform retardation. The tangenl to the curve A C 
at the point E will make an angle of 30° with E F, and D F will 
be the pitch line. 

Eight construction points were taken in developing the curve 
A C. Eight points will be sufficient for beginners for practice work 




Fig. 20. — Elliptical Curve 



and later six points may be used. When using six points only nine 
equal divisions should be laid out on the line R S, the remaining 
construction being the same as described above, except that D E 
should be divided into three parts instead of four. In practical work 
many more construction points should be used for accuracy and 
smooth cam action. 

37. Elliptical curve. Fig. 26. Draw rectangle A B C R. 



Draw semi-ellipse making F G equal to -r F C. 



To draw the ellipse, 



take a strip of paper with a straight edge and mark fine lines at 
P, T, and S, Fig. 26a, making P T = C F and P S = G F. Move 
the strip of paper so that S will always be on 
the line R C, and T on the line F G; P will then 
describe the path of the ellipse. Having the semi- 
ellipse, divide the part R G, Fig. 26, into four 
equal arcs as at 1, 2, 3. This is quickest done 
by setting the small dividers to a small space of 
any value and stepping off the distance from R 
to G. Suppose that there are 18.8 steps. Set 
down this number and divide it into four parts, 
giving 4.7, 9.4, and 14.1. Then again step off the 
arc from R to G with the same setting of the dividers, marking the 
points that are at 4.7, 9.4, and 14.1 steps. The compass setting- 
being small, the fractional part of it can be estimated with all prac- 
tical precision. Divide D E into four equal parts as at H, I, J. 
Draw vertical lines from these points and horizontal lines from the 




Fig. 2Ga. — Showing 
Method of Draw- 
ing Semi-Ei.lip.se 



24 ELEMENTARY CAMS 

corresponding points at 1, 2, and 3. The intersections, as at K, 
will give a series of points on the elliptical base curve. The curve 
E C is similar to A E but in reverse order. The tangent to the curve 
at E makes an angle of 30° with E F, and D F is the pitch line. 

The elliptical base curve gives slower starting and stopping 
velocities to the follower than any of the other curves, but the velocity 
is higher at the center of the stroke. The acceleration is variable and 
increases to the middle of the stroke, where its maximum value is 
greater than that of the crank curve but less than that of the straight- 
line combination curve. The retardation values decrease in reverse 
order to the end of. the stroke. 



SECTION III.— CAM PROBLEMS AND EXERCISE 
PROBLEMS 

38. Problem 1. Empirical design. Required a radial cam 
that will operate a V-edge follower: 

(a) Up 3 units while the cam turns 90°. 

(b) Down 2 " " " " " 60°. 

(c) Dwell " " " " 120°. 

(d) Down 1 unit " " " " 90°. 

39. Applying the simplest process for laying out cams, it is only 
necessary, in starting, to assume a minimum radius C D, Fig. 27, for 




Fig. 27. — Empirical Design of Cam for Data in Problem 1, V-Edge Follower 

the cam, and then lay off the given or total distance of 3 units as 
at D B. The assigned angle of 90° is next laid off as at D C Di and 
the point Di marked so as to be 3 units further out than D. Any 
desired curve is then drawn through the points I) and l) x and part 
of the cam layout is completed. The same operations are repeated 
for obtaining the points D 2 and D 3 and the entire cam is finished. 
If the follower had roller contact instead of V-edge contact, a 

25 



26 



ELEMENTARY CAMS 



minimum radius C D, Fig. 28, would be assumed as in the previous 
case, and D would be taken as the center of the roller. The closed 
curve D, D 4 , Z>i . . . would be obtained as before and another closed 
curve E, Ei . . . would be drawn parallel to it at a distance equal 




Fig. 28 — Empirical Design of Cam for Data in Problem 1, Roller Follower 



to the assumed radius of the roller. The latter closed curve would 
be the actual outline of the cam. 

The closed curve E E\ . . . would be known as the working sur- 
face and the curve 2) Di ... as the pitch surface of the cam. In 
Fig. 27 the pitch and working surfaces coincide because the follower 
has a V-edge. 

40. Cams are sometimes designed with no more labor than that 
entailed in the previous preliminary problem. And it may be added 
that where one has had a sufficient experience good practical results 
may be obtained by following only this simple method. 

The method of cam construction described above, however, does 
not enable the cam builder or designer to hold in control the velocity 
or acceleration of the follower rod D G as it moves up its 3 units; 
nor does it enable him to know the variable and maximum side pres- 
sures which exist between the follower rod and the bearing or guide 



CAM PROBLEMS AND EXERCISE PROBLEMS 



27 



F, Fig. 27, as the rod moves up. In order that these things may 
be known, this preliminary problem will now be redrawn with ad- 
ditional specifications. 

41. Problem 2. Technical desicn. Required a radial cam that 
will operate a roller follower: 

(a) Up 3 units while the cam turns 90°. 

(b) Down 2 " " " " " 60°. 

(c) Dwell " " " " 120°. 

(d) Down 1 unit " " " " 90°. 

(e) The follower, in all its motions, shall move with uniform 
acceleration and uniform retardation. 

(f) The maximum side pressure of the cam against the follower 
rod shall be 40°. 

Items (a), (b), (c), and (d) are the same as in Problem 1. 

42. Inasmuch as this problem is given at this place simply to 
show that velocity and acceleration and side pressure can always 
be controlled with very little additional labor beyond that necessary 
for the simple layout shown in Fig. 28, the full explanations of the 
•formula and figures used will not be given here. They will be taken 
up in their proper order in subsequent paragraphs. For this problem 
the only necessary computation is : 



57.3 



hf 



57.3 



3 X 2.38 



b "'•" 90 

C H, Fig. 29. 

The reference letters, h, f, 
and b are defined in paragraph 
29. Lay off C H in Fig. 29, and 
then lay off the- follower motion 
of 3 units equally distributed on 
each side of H, as at H B and 
H D. Divide D H into nine 
equal parts and take the first, 
fourth, and ninth parts; do like- 
wise with B H. Divide the 90° 
angle B C Dx into six equal 
parts by radial lines as shown, 
and swing each of the six di- 
vision points between D and B 



= 4.55 = Radius of pitch circle = 




around until they meet succes- FlG - 29 — technical design of cam for data 

in Problem 2, Drawn to Same Scale 
s - as Fia. 28 



sively the six radial line 



28 



ELEMENTARY CAMS 




A curve through the intersecting points will be the pitch surface of 
the cam, as shown by the dash-and-dot curve D Hi D\. . . . 

The working surface will be E Ei 
. . . which is found as described 
in paragraph 26. 

The pitch surface Z>i Z) 2 is 
obtained in the same way as 
D Di was found. The curve 
D 2 Ds is an arc of a circle, and 
the curve D 3 D is found in the 
same manner as D D\. 

43. Advantages of the 

TECHNICAL DESIGN. With the 

cam constructed as above the 
follower will start to move with 
the same characteristic motion 
as has a falling body starting 

Fig. 29.— (Duplicate) Technical Design of f rom rest anc J t } ie follower will 
Cam for Data in Problem 2, Drawn to . 

same Scale as fig. 28 be stopped with the same gen- 

tle motion in reverse order. It- 
will be definitely known also that the greatest side pressure 
of the cam against the follower is at an angle of 40° as specified, 
and that this pressure will occur when Hi of the pitch surface of the 
cam is at H, or when the roller is in contact with the working 
surface at H 2 . Where the cam form is assumed as in Fig. 28, nothing 
is known positively of the starting and stopping velocities of the 
follower. Further, as may be found by trial, the maximum angle of 
pressure of the cam against the rod runs up to 47° in Fig. 28, as 
shown at D 4 . The minimum radius of the cam in Fig. 28 was taken 
equal to that in Fig. 29 for comparison. 

44. The two previous problems have been given as brief exercises 
without going into all the detail necessary to a full understanding, 
in order to give an idea of the method of producing cams on a scientific 
basis. In the problems which will follow, the several steps in building 
cams of various types will be explained. In many of the problems 
the same data will be used so that comparisons of different forms 
of cams which produce the same results may be made. 

45. Problem 3. Single-step radial cam, pressure angle 
equal on both strokes. Required a single-step radial cam in 
which the center of the follower roller moves in a radial line. The 
maximum pressure angle to be 30°, and the follower to move: 



CAM PROBLEMS AND EXERCISE PROBLEMS 



29 





(a) Up 3 units in 90° with uniform acceleration 
and retardation. 

(b) Down 3 units in 90° with uniform 
acceleration and retardation. 

(c) At rest for 180° 

46. The first step in the solution is to determine 

the total length of the cam chart for a parabola 

chart curve and for a 30° maximum pressure angle. 

From the table, paragraph 30, the factor for this 

case is found to be 3.46. Since the travel of the 

follower is 3 units in 34 revolution, the total length 

of chart will be 3 X 3.46 X 4 = 41.52, which, 

therefore, is the length of the chart A A' in Fig. 30. 

This length represents the 360° of the cam. Lay 

off A W equal to 90°, according to item (a) in 

the data. Construct the parabolic curve A E C. 

Completing the entire chart, the base curve is 

found tobe AC M N A'. The next step is to find 

the radius of the pitch circle. The circumference 

of this circle is equal to the length of the pitch . 

41.52 
line D D f . Its radius is, therefore, equal to ~ — = 

Z IT 

6.61, and this value is laid off at D, Fig. 31, and 
the pitch circle D F Q W drawn. The quadrant 
D F is divided into the same number of parts 
as D F in Fig. 30. The vertical construction lines 
H Hi, II i, J J i . . . in Fig. 30 now become the 
radial lines correspondingly lettered in Fig. 31, 
and the pitch surface is drawn through the points 
A Hi 7i J\. . . . The positions of maximum pres- 
sure are shown at E and Q; at all other points it 
will be less. The working surface B G R P is 
found by assuming a radius A B for the roller, 
and by striking a series of arcs as shown at H 2 , 
L, J 2 . . • with the points H h I h Ji ... as cen- 
ters, and then drawing the working curve tangent 
to these arcs. With the same specifications for 
the up and down motions of the follower, as 
given by items (a) and (b) in the data, this type 
of cam will be symmetrical about the line Y C. 



30 



ELEMENTARY CAMS 




Fig. 31. — Problem 3, Cam Laid Out from Cam Chart 




Fig. 32. — Problem 3, Cam Laid Out Independently of Cam Chart 



CAM PROBLEMS AND EXERCISE PROBLEMS 31 

47. Omission of cam chart. When the relation between pics- 
sure angle, chart base and pitch lines, and cam pitch and surf; ice 
lines is understood and fixed in mind, the actual drawing of the 
chart for the graphical construction of simple cams and particularly 
of single-step cams may be omitted with full confidence when the 
elementary base curves are used. For example, the problem in the 
previous paragraph is shown completely worked out in Fig. 32 
without any reference whatever to the chart of Fig. 30. The radius 
D of the pitch circle, Fig. 32, is obtained directly from the formula, 

h f 

r = 57.3 -j- given in paragraph 29. Substituting the data as given 

3 X 3.46 
in the previous paragraph, r = 57.3 ^ — = 6.61 and is laid off 

at D 0. The assigned motion of the follower is laid off symmetri- 
cally on both sides of the pitch point D, as at A V, and the distances 
A D and V D are divided into the desired number of unequal parts, 
as at 1, 4, 9, 16. The quadrant D F is divided into the same number 
of equal parts as at H, I, J . . . and indefinite radial construction 
lines drawn through the points. Circular construction arcs are 
next drawn through the points 1, 4, 9 . . . until they intersect the 
radial lines, thus obtaining points H h I h J x . . . on the cam pitch 
surface. In general, a neater construction is obtained by omitting 
the full length of the construction arcs, as from V to C . . . and 
simply drawing short portions of the arc at the intersecting radial 
lines as shown in the lower left-hand quadrant between C and M . 

48. Exercise problem 3a. Required a single-step radial cam 
in which the center of the follower roller moves in a radial line. 
The maximum pressure angle to be 40°, and the follower to move: 

(a) Out 6 units in 135° on the crank curve. 

(b) In 6 " " 135° " " " 

(c) At rest for 90°. 

49. Problem 4. Single-step radial cam, pressure angles 
unequal on the two strokes. Required a single-step radial cam 
in which the center of the follower moves in a radial line. The 
maximum pressure angle not to exceed 30° on the outstroke nor 50° 
on the return stroke, and the follower to move: 

(a) Out 2 units in /i 6 revolution on the crank curve. 

(b) In 2 " " % 6 " " " 

(c) At rest for y 2 revolution. 



32 



ELEMENTARY CAMS 



50. The diameter of pitch circle of the cam that will be necessary 
to fulfil the requirements on the outstroke will be: 

2 X 2.72 X 16 
d a = — » * - — = 5.54 units, or from formula paragraph 29. 

r = .159 2X2 - 7 5 2X16 = 2.77, 

and the diameter of pitch circle required for the instroke will be 

, 2 X 1.32 X 16 ... . 
b = 3 14 X 3 — = units. 

Inasmuch as there can be only one pitch circle for a cam, the 
largest one resulting from the several specifications must be used. 
In this problem then the diameter S D of the pitch circle in Fig. 33 




Fig. 33. — Problem 4, Maximum Pressure Angle Different on the Two Strokes 



equals 5.54 units. The follower's motion of two units is laid out 
at A V and the pitch surface A E C M N constructed. The working 
surface of the cam B KG, etc., is then drawn. Since a larger diameter 
of pitch circle had to be used for the return stroke than the require- 



CAM PROBLEMS AND EXERCISE PROBLEMS 33 

merits called for, it follows that the pressure angle will not reach 
50° on that stroke, and it may be of some interest to determine what 
the maximum pressure angle on the return stroke will be. Sub- 

7 f 

stituting the diameter used, 5.54, in the formula d = — and solving 

for/, / is found to be equal to 1.63. From the chart in Fig. 21 it 
is shown that a factor of 1.63 for the crank curve corresponds to a 
maximum pressure angle of nearly 44°, and this angle may be drawn 
in its proper position at Q in Fig. 33. 

51. Exercise problem 4a. Required a single-step radial cam 
in which the center of the follower roller moves in a radial line. The 
maximum pressure not to exceed 30° on the up stroke nor 40° on the 
down stroke, and the follower to move: 

(a) Up 3 units in 135° on the parabola curve. 

(b) At rest for 45°. 

(c) Down 3 units in 90° on the parabola curve. 

(d) At rest for 90°. 

52. Pressure angle increases as pitch size of cam decreases. 
This is illustrated in Fig. 34, where the large pitch cam represented 
by D, D 2 . . . gives exactly the same motion to a follower as the 
small pitch cam d, d 2 . . . . It will be noted that the pressure angle 
for the large cam, at the start, is H D G, while for the small cam it 
is increased to h d g. Likewise the maximum pressure angle for the 
large cam, when the follower is near the end of its stroke, is &i, 
while for the small cam the maximum pressure angle is b, which is 
larger than b u From these observations it may be said, in general, 
that the larger the pitch surface of the cam the smaller will be the 
pressure angle. The size of the roller has no effect whatever on 
the pressure angle. Two cams of the same pitch size may be of 
totally different actual sizes for the same work, one cam having 
a large roller and the other a small roller. Therefore it is important 
to remember that, in general, the pressure angle may be regulated 
by changing the size of the pitch surface only and not the working 
surface. 

53. Change of pressure angle in passing from chart to 
cam. The circumference of the pitch circle of the cam, it will be 
recalled, is equal to the length of the pitch line on the chart. It 
will also be remembered that the pitch line may be at various heights 
on the chart, paragraph 23. It is now important to consider: 

1st. That the pressure angle at the pitch circle on the cam must 
be the same as the pressure angle at the pitch line on the chart. 



34 



ELEMENTARY CAMS 



2d. That the pressure angle at any point on the pitch surface 
of the cam outside of the pitch circle will be less than the pressure 
angle of the corresponding point on the base curve of the cam chart. 

3d. That the pressure angle at any point on the pitch surface 




d^ — — — _ x> 5 

Fig. 34. — Showing Relation Between Pressure Angle and Size of Pitch Cam 

of the cam inside of the pitch circle will be greater than the pressure 
angle of the corresponding point on the base curve of the cam chart. 

These statements, which are theoretically true for nearly all cases, 
and practically so for all other cases where the usual base curves 
are employed, are demonstrated in the following paragraph. 

54. Cam considered as a bent chart. Consider that the cam 
itself is the cam chart bent in its own plane so that the pitch line 



CAM PROBLEMS AND EXERCISE PROBLEMS 



35 



becomes the pitch circle. Then the line D D' ', Fig. 30, becomes 
the circle DFOW, Fig. 31; the line V V f is stretched to become 
the circle V C S Y, and the straight line A MA' is compressed to 
become the circle A MA. This means, in a general way, that 
the rectangle D V V f D', Fig. 30, is so distorted that if an original 
diagonal had been drawn from D to V' it would have an increased 
length and a decreasing slant after the bending had taken place. 
With a decreasing slant of the pitch surface the pressure angle will 
decrease. Likewise, a diagonal drawn from D r to A in the original 
rectangular chart would be decreased in length and would have an 
increasing slant, and the pressure angle would be increasing toward 
A. This is illustrated in detail in Figs. 35 and 30. 

55. Base line angles, before and after bending. The pres- 
sure angle of 30° at E in Fig. 35 is reduced to 23° in Fig. 30, and the 




Fia. 35. — Section of Cam Chart Be- Fig. 36. — Section of Cam Chart After 
fore Bending Bending, BC Constant in Both Figubbs 



30° at D are increased to 41°. Fig. 35 represents a cam chart with 
a straight base line D E, and Fig. 30 is a corresponding cam sector 
with D E as the pitch surface. If B C, Fig. 35, is taken as the 
pitch line, B C, Fig. 30, will be part of the pitch circle. The uniform 
pressure angle of 30° from A to E, Fig. 35, will grow smaller beyond 
A in Fig. 30 for the reason that the radial components of the tan- 
gential triangles remain constant, as illustrated at L M, while the 
tangential components grow longer as illustrated from A N to E L, 
which are respectively equal to the arcs A Y and E L\. Con- 
sequently, the angles grow smaller from the angle N A P to L E M. 
Similarly it may be shown that they grow larger from NAP to 
QDR. 

50. Limiting size of follower roller. The radius of the 
follower roller may be equal to, but in general should be less than 



36 



ELEMENTARY CAMS 



the shortest radius of curvature of the pitch surface, when measured 
on the working-surface side. If the radius of the roller is not so 
taken, the follower, when put in service, will not have the motion 
for which it was designed. 

57. Case 1. Radius of roller equal to radius of curvature 
of pitch cam. In Fig. 37, A B E F A is the pitch surface of a cam. 




Fig. 37. — Limiting Size of Follower Roller 



CAM PROBLEMS AND EXERCISE PROBLEMS 37 

G A is the radius of curvature at A and A G is the radius of the 
roller. In this case both radii are equal and the working surface 
has a sharp edge at G. 

58. Case II. Radius of roller greater than radius of 
curvature of pitch cam. From B to C, Fig. 37, the radius of 
curvature of the pitch surface is H B, which is less than the roller 
radius. In this case the working surface will be undercut at / in 
generating the cam, and if the cam is built the center of the roller 
will mark the path BiJiCi instead of Bi J Cj, and the follower will 
fail to move the desired distance by the amount J J±. 

59. Special application of case II. Effect of an angle in 
the pitch surface outline. This is illustrated at R F Q in Fig. 37, 
and is a special application of Case II, in w-hich the radius of cur- 
vature of the cam's pitch surface is reduced to zero. Undercutting 
is here illustrated by considering that a cutter, represented by the 
dash circular arc, is moving with its center on the pitch surface 
arc E F. It then cuts the working surface M S. As the center of 
the cutter is moved from F toward A, the part W S of the working 
surface which was previously formed is now cut away, leaving 
the sharp edge W on which the follower roller will turn when the 
cam is placed in operation. The center of the follower roller will 
then move in the path R T Q instead of R F Q, and the follower will 
fall short of the desired motion by the amount T F. 

60. Case III. Radius of roller less than radius of curva- 
ture of pitch cam. From D to E, Fig. 37, the radius of curvature 
of the pitch surface is K D, which is greater than the roller radius. 
In this case, which is the practical one, although close to the limit, 
a smooth curved working surface is provided for the roller from 
Lto M. 

61. Radius of roller not affected by radius of curvature 
on non-working side. From Ci to D, Fig. 37, the radius of curva- 
ture of the pitch surface is less than the radius of the roller, but 
this short radius is not on the working side of the pitch surface, 
and therefore the roller will roll on the surface I L while its center 
travels on the pitch curve Ci D. 

62. Rollers for positive-drive cams. When the largest roller 
for a positive or double-acting cam is being determined the radius 
of curvature on both sides the pitch-surface curve must be con- 
sidered and the smallest radius used. For example, in Fig. 37, if 
A J E T A were the pitch surface for a double-acting cam, N C 
would be the maximum roller radius, whereas H J would be 



38 



ELEMENTARY CAMS 



the maximum radius if it were for an external single-acting 
cam. 

63. Radius of curvature of non-circular arcs. In illustrat- 
ing the above cases the pitch surface was assumed as being made 
up of straight lines and arcs of circles in order to show more effec- 
tively and more simply the limits of action in each instance. Where 
the pitch surface contains curves of constantly varying curvature, 
and they generally do in practice, the shortest radius of curvature 
of the pitch surface may be found with all necessary accuracy by 
trial with the compass, using finally that radius whose circular arc 
agrees for a small distance with the irregularly curved arc. For 
example, in Fig. 38, let G H D J B be a portion of a pitch surface 




Fig. 38. — Limiting Size of Follower Roller Working on Non-Circular Cam Curves 



made up of non-circular arcs. The shortest radius of curvature 
on both sides is found, by trial, to be F H. The center F is marked 
and the osculatory arc X H Z drawn in. Then H F is the largest 
possible radius of roller for a double-acting cam, and with this 
roller the working surfaces will be V F T W and Vi Fi T x W\. 



CAM PROBLEMS AND EXERCISE PROBLEMS 30 

If a larger roller is used, with a radius I) I\, for example, the 
working; surfaces of the groove will be S E and l\ A'i A',, and the 
new pitch surface, after cutting the cam, will be GC DLB, if the 
roller is kept always in contact with the inner surface of the groove. 
If it is kept always in contact with the outer surface of the groove, 
the original pitch surface will be changed to G C H DiJ B. In 
either case the original desired follower motion is not obtained if the 
roller is too large, and if a positive-drive cam is run with the larger 
roller the follower's motion will be indeterminate, the center of 
the roller having any possible position between C D L and C H J L. 

64. Problem 5. Double-step radial cam. Required a double- 
step radial cam in which the center of the follower roller moves in 
a radial line. The maximum pressure angle to be 30°, and the 
follower to move: 

(a) Up 4 units in }/$ revolution on the crank curve. 

(b) At rest for 34 revolution. 

(c) Up 4 units in y% revolution on the parabola curve. 

(d) Down 2 units in % revolution on the elliptical curve. 

(e) At rest for Y% revolution. 

(f) Down 6 units in \i revolution on the parabola curve. 

65. In Problem 3 there are only two motion assignments, (a) and 
(b), in the data, and they were the same except for direction. Con- 
sequently only one computation was necessary. When two or more 
dissimilar assignments are made in the data, as in the present problem, 
it is advisable to make a computation for the length of the chart 
diagram for each motion specification, as follows: 

(a) 4 X 2.72 X 8 = 87.04, which is the length of chart and of 

the pitch circle circumference = 
13.86 pitch circle radius. 

(c) 4 X 3.46 X 8 = 110.72, which is the length of chart and pitch 

circle circumference = 17.62 pitch 
circle radius. 

(d) 2 X 3.95 X 8 = 63.20, which is the length of chart and pitch 

circle circumference = 10.06 pitch 
circle radius. 
(f) 6 X 3.46 X 4 = 83.04, which is the length of chart and pitch 

circle circumference = 13.22 pitch 

circle radius. 

Inasmuch as there is a different length of chart and a different 

pitch line for each item in the data one can not tell which pitch line 

to take without some preliminary computation. For this purpose 



40 



ELEMENTARY CAMS 



a chart diagram is well adapted, as follows: Construct a rectangle, 
Fig. 39, with a height A T equal to the total motion of the follower 
in one direction, 8 units in this case. Make the length A A' of 
rectangle any convenient value entirely independent of any of the 
values computed above and label this according to the longest chart 
length as computed above. Lay off straight lines to represent the 
component parts of the base curve as assigned in the data and label 
them as shown at A C, C B, B H, etc. Draw the several pitch lines 
as at F D, J I, etc. 

66. For general procedure, consider the pitch line which passes 
through the point calling for the longest chart length. This will 
be the pitch line J I passing through G, Fig. 39, which calls for a 
chart length of 110.72 and a pitch radius of 17.62. If G is to be at 



T 










H 














* 


u 

D 






~u I 


.$*y 


n^ 










-Lj 


, 


yr* j 


vl 


i\t 


U 






M, f L\ 




B 





1 


S 






f 


C 4 


Q 


si 


*X 












f F 


> ■ 


<y 


P 


t 

CO 

1 






* 


eo 

•I ■ 


A 






£\ 




% 9 




X 




























' 





Fig. 39. — Problem 5, Cam Chart Diagram for Double-Step Cam 



a radius of 17.62 in the cam, E will be at a radius of 17.62 - 4 = 
13.62. But from computation (a) it is seen that E must be at a 
radius of 13.86. Therefore, if the radius of cam pitch circle is 
retained at 17.62, the trial pitch line, J I, on the chart diagram will 
have to be lowered, 13.86 - 13.62 = .24, giving the new pitch line 
U U'. If the line U U' now becomes the pitch circle the point E 
will be at 17.62 - 3.76 = 13.86, just as called for in computation 
(a), and the pressure angle will be 30° at the point E on the cam. 

The other critical points at P and K must also be tested with 
respect to the proposed pitch line, U U'. With this pitch line the 
point P will be 2.76 inside of the pitch circle, or at a radius of 17.62 — 
2.76 = 14.86. This is safe, as the computed radius for P was only 
13.22 according to item (f). The point K is also safe, for it will be 
at a radius of 17.62 + 1.24 = 17.86, whereas a radius of only 10.06 
is required. 

67. The cam may now be drawn by constructing the true cam 
chart as in Fig. 40, which is lettered the same as Fig. 39, and plotting 
the cam from it as in Fig. 41. The pitch line U U' of Fig. 40 be- 



CAM PROBLEMS AND EXERCISE PROBLEMS 



41 



&j ^ 



w 



^3 



&s b o» ^ H 



comes the pitch circle, having a radius U in 
Fig. 41, and the ordinal cs of Fig. 40 become 
the radial measuring lines in Fig. 41. Or the 
cam may be drawn directly, without the use of a 
cam chart, as indicated in Fig. 42, where the 
pitch circle V is first drawn with a radius of 
17.62. The assigned angles are then laid down 
and the several pitch curves, such as A' E C, are 
constructed at the proper radial distances as de- 
termined in Fig. 39 and as illustrated for one 
case at E Ei (3.76) in Fig. 42. 

68. Determination of maximum pressure 
angle for each of the curves making up a 
multiple-step cam. If it is desired to know the 
exact pressure angle at P, Fig. 41, it may be readily 
determined by making the value of r = (17.62 — 



14.86 in the formula, r = .159 and 

e 



2.76) 

solving for /, the notation being the same as given 
in paragraph 29. 

14.86 



/ = 



159 X 6 X 4 



= 3.89. 



Consulting the chart of cam factors in Fig. 21, 
it is found that a factor of 3.89, when applied to 
the parabola chart curve, shows a cam pressure 
angle of about 27°, which is under the assigned 
limit, and therefore need not be further consid- 
ered. In a similar manner the pressure angle at 
G and K on the cam may be computed if desired, 
the former being a small fraction of a degree 
under 30° and the latter something less than 20°, 
the reading running off the chart. 

69. Exercise problem 5a. Required a double- 
step radial periphery cam in which the center of 
the follower roller moves in a radial line. The 
maximum pressure angle to be 30° and the follower 
to move: 

(a) Out 5 units in 150°, with uniform accelera- 
tion and retardation. 

(b) In 2 units in 45° on the crank curve. 



42 ELEMENTARY CAMS 

(c) At rest for 45°. 

(d) In 3 units in 120° on the crank curve. 

70. Problem 6. Cam with offset roller follower. Re- 
quired a single-step radial periphery cam in which the center of the 
follower roller moves forth and back in a straight line which does 
not pass through the center of rotation of the cam. The maximum 
pressure angle when the follower is at the bottom of its stroke is 
to be 30°, and the follower is to move: 

(a) Up 3 units in 90° on the parabola curve. 

(b) Down 3 " " 90° " " 

(c) At rest for 180°. 

71. Problems of this nature are totally different, both in pressure- 
angle action and in methods of construction, from the preceding 
ones. As may be noted in the data, it is required that the pressure 
angle, when the follower is at rest at the bottom of its stroke, shall be 30°. 
It will appear presently that the pressure angle, when the follower 
is in motion, may be zero or even negative on one of the strokes in 
this form of cam. It will also be shown that the maximum pressure 
angle during the follower motion cannot be assigned in advance and 
obtained in any practical manner. From the above it follows that 
the offset radial cam has a peculiar advantage in keeping considerable 
side pressure off the follower guides during the time that the follower 
is moving in one direction, although at the bottom of the stroke the 
pressure angle may have any desired value, and during the period 
of motion in the opposite direction the pressure angle will reach a 
maximum value much larger than the assigned angle at the bottom 
of the stroke. 

72. The method of construction for the offset roller cam is illus- 
trated in Fig. 43. The diameter of the pitch circle, U F r S T, is com- 

puted as before by the formula, d = 114.6 -j~, and found to be 13.22 

units. An angle equal to the assigned pressure angle is then laid 
off at U U , U being parallel to the direction of motion of the 
follower. Draw a line, DW, parallel to U and so located that it 
has an intercept D A between the pitch circle and the inclined line 
equal to one-half the travel of the follower. This may be done by 
trial, or graphically, as shown by the dotted-line construction which 
is drawn at Y' X A' instead of at Y to avoid complication of con- 
struction lines. The angle U O Y' equals the angle U O Y. Y r X, 
parallel to U 0, is drawn equal to one-half the stroke. An arc X A\ 



CAM PROBLEMS AND EXERCISE PROBLEMS 



\:\ 



parallel to the arc Y' U, is drawn through X by using U as a radius 
and Z as a center, where Z equals )'' X. A circular arc through 
A' with as a center will intersect U r in the desired points. The 
point A will then be the lowest point of the stroke, I) will he the 
center of the stroke, and \Y the radius of the construction circle. 




Fig. 41. — Problem 5, Double-Step Cam Constructed from Cam Chart 



U' Mo 




Fig. 42. — Problem 5, Double-Step Cam Constructed Without Use of Cam Chart 



The distance A V is equal to the assigned 3 units of motion, and 
the divisions 1, 4, 9 . . . are made according to the requirements of 
the parabola curve. The assigned 90° is laid off on the construction 
circle at W F and divided into a number of equal arcs at H, I, 
J . . . corresponding to the number of divisions at A V, eight being 
used in the present example. Tangents to the construction circle, 



44 ELEMENTARY CAMS 

such as H H h I I h J Ji . . . are then drawn at H, I, J . . . and the 
distances Wl, W4, W9 . . . laid off on these tangents, thus giving 
the points H h I h Ji . . . on the pitch surface of the cam. Or these 
latter points may be obtained by swinging arcs through 1, 4, 9 . . . 
about as a center, until they meet the respective tangents at H h 

73. An examination of the pressure angles for a cam with an 
offset follower shows that during the up stroke the pressure angles 
are very small, being, in fact, negative from J\ to K h Fig. 43, and, 
when measured, the average pressure angle for the working or up 
stroke is between 6 and 7 degrees in this problem; although on the 
down or return stroke it reaches an average of between 37° and 38° 
and a maximum of 46° near Q f . In this class of problem the com- 
putation for diameter of pitch circle serves merely as a guide in deter- 
mining a size that will give a small cam and a small average pressure 
angle on the working stroke. If the diameter of the pitch circle is 
arbitrarily taken either larger or smaller than the value, as above 
computed, or if other base curves are used, the negative pressure 
angles at J h E h and Ki may disappear entirely; which would be an 
advantage where it is desired to have pressure on the follower guides 
on one side only. 

74. It has doubtless been observed that there is a decided lack 
of symmetry in this form of cam, even though the data are similar 
for both strokes of the follower. This is illustrated in Fig. 43, where 
the portion A C of the pitch surface for the outstroke is quite different 
from the portion C M. It is also characteristic of this form of cam 
that the pitch and working curves each embrace either a smaller or 
a larger angle than the assigned angle for a given stroke of the follower, 
as shown by the angle A C being less, and the angle COM being 
greater, than the assigned 90°. This, of course, is due to the fact 
that when C has traveled 90° to V the line C will have passed the 
original zero line A of the pitch curve and will be in the position 
V. Therefore, the cam angle for one stroke of the follower will be 
less than the assigned angle by the amount of the angle included by 
V A; for the other stroke it will be greater than the assigned 
angle by the same amount. 

75. Exercise problem 6a. Required a single-step radial periph- 
ery cam in which the center of the follower roller moves forth and 
back in a straight line which does not pass through the center of 
rotation of the cam. The maximum pressure angle when the follower 
is at the bottom of its stroke is to be 30°, and the follower is to move: 



CAM PROBLEMS AND EXERCISE PROBLEMS 45 



(a) Out 6 units in 135° on the crank curve. 

(b) In 6 " " 135° " " " 

(c) Rest for 90°. . 

In this problem, only the initial pressure angle at the bottom of 
the stroke need be shown; the pressure angles at other positions, 
such as are shown in Fig. 43 at Hi, I h may be omitted. 



Fig. 43. — Problem 6, Cam with Offset Roller Follower 

76. Problem 7. Cam with flat surface follower, — Mush- 
room cam. Required a radial periphery cam to operate an offset 
follower which has a flat surface instead of a roller. The follower 
to move: 

(a) Up 3 units in 90° on the parabola base. 

(b) Down 3 " " 90° " " 

(c) At rest for 180°. 

77. This type of cam is known also as the mushroom cam. Flat 



46 



ELEMENTARY CAMS 



surface followers may be offset as shown in the side and top views 
in Fig. 44, where the center line N" Y" of the follower spindle is set 
the distance P" 0" in front of the center of the cam plate. In this 
case there will be a part sliding and part rolling of the cam on the 
follower and the follower will turn about its own axis, N" Y," as it 




Fig. 44. — Problem 7, Cam with Flat Surface Follower — Mushroom Cam 



is being raised and lowered. When the follower is not offset, i.e., 
when the center line 0" N" is placed in line with M" P", the action 
will be all sliding and there will be no turning of the follower spindle 
on its axis. In this case there will be localized wear on the follower, 
while in the former case the wear will be more widely distributed 
over the follower surface. In both cases the construction is the 
same and is explained in the following paragraph. 

78. In cam followers having flat surfaces perpendicular to the 
line of action, the line of pressure is M " Q" and is parallel to the 
line of action of the follower, instead of being inclined to it as in 
the case of cams having roller followers. Because of this character- 
istic action the ordinary pressure-angle factors do not apply in 



CAM PROBLEMS AND EXERCISE PROBLEMS 47 

cams of this class in computing or obtaining the diameter of the 
pitch circle D F S T, and this circle may be assumed. In some 
cases a fair guide for the size of this circle may be obtained by using 

7 f 

the regular formula, d = 114.6 -r, for diameter of pilch circle, as- 
suming the 30° pressure angle factor. Solving, r is found to equal 
6.61, and is laid off at D. The assigned three units of motion are 
then laid off, one-half on each side of D, as at A and V. The assigned 
90° are next laid off at A F and divided into the desired number 
of construction parts, four being used in this case, as at D\ y D 2 , 
. . . The distance A V is also divided into four parts, A H and V K 
being each equal to 1 unit and H D and K D equal to 3 units. Only 
four divisions are taken in this case to avoid confusion of lines in 
the illustration, but in student problems 6 or 8 points should be 
taken, and in practical work 12 to 24 divisions should be used. 
The first division point, H, is now revolved to meet the first radial 
division line D h thus giving the point H h where a line Hi E is 
drawn perpendicular to Hi 0. This line Hi E represents the bottom 
of the follower disk A C with reference to the cam when the cam 
has turned through the angle A Hi. The points D 2 and Ki are 
obtained in the same manner as was Hi and corresponding perpendic- 
ulars are drawn, as at D 2 Ai and Ki K 2 . As smooth a curve as possible 
is now drawn tangent to these perpendiculars and the points of 
tangency marked as at H 2 , D 4 , and K 2 . This smooth curve, A G, 
is the working surface of the cam. 

79. The size of the follower must also be determined. The most 
satisfactory way of doing this is to find, first, the locus, or path, 
of the line of contact between the periphery of the cam and the 
follower disk. This is obtained by considering that when Hi is at 
H, the point of tangency H 2 is at Hz, the length H H 3 being equal to 
Hi H 2 . Likewise, when D 2 is at D, D 4 is at D , and the same for the 
other points of tangency. The dash line curve through the points 
A Hs D 5 K 3 . . . is the locus of contact between the cam and the 
follower. The point L is the extreme point of this curve and if the 
follower were not offset, the length of an ordinary toe or flat extension 
of the follower would have to be at least equal to N f X. f If the 
follower is offset, say by the amount A/' M' (= N" M") } the radius 
of the disk will have to be at least equal to N f Z/, and the extreme 
line of contact will be L f J r . The other extreme line of contact will 
be a similar line through L ,ff , and the area of the flat disk which will 



48 



ELEMENTARY CAMS 



be subject to wear will be the annular surface between the periphery 
and the dashline circle whose radius is N f A' . As to the wear on the 
cam itself, there would be pure sliding of the curved surface A G 
on the flat surface A X if the follower were not offset. With an 
offset follower there is an effective turning radius equal to the offset 




Fig. 44. — (Duplicate). 



Problem 7, Cam with Flat Surface Follower- 
Mushroom Cam 



N' M r tending to rotate the follower about its axis N" Y", and this 
changes the action of the cam on the follower entirely by causing 
part rolling and part sliding. 

80. The pressure angle in this form of cam must be considered 
differently from cams which operate against rollers. In roller 
follower cams it is the angle between the normal to the cam surface 
and the line of action of the follower that determines the side pres- 
sure on the bearings, whereas in flat surface followers it is the distance 
that the line of contact is away from the line of action that determines 
it. This distance varies constantly, and in the illustration in Fig. 
44 the limits of variation are M' N r and Q' N'. These are, in reality, 
lever arms on which the pressure acts to produce a turning moment 



CAM PROBLEMS AND EXERCISE PROBLEMS 49 

which must be resisted by the follower guides. Since there can be 
no pure rolling action between the cam and follower in constructions 
of this type, there is nothing to be gained in this particular by a 
large offset. On the contrary, there is much to be lost, due to the 
large turning moment on the follower rod. A fair guide as to the 
offset would be to keep the angle formed by the center line Y" 0" 
of the follower motion and the line Y" M" or Y" Q" joining the 
center of the bearing with the midpoint of the line of contact, to 
within, say, 30°, or any other maximum value that circumstance 
might warrant. The angle here defined might be termed the pres- 
sure angle in this type of cam. The minimum pressure angle, 
N" Y" M", is seen in its true size, while the maximum pressure 
angle as projected at U" Y" Q" must be revolved about U" Y" 
as an axis until U" Q" equals N' Q f , when it will appear in its true 
size as at U" Y" Q"' . 

81. Limited use of cams with flat surface followers. Cams 
with followers of this type are not well adapted, in general, for cases 
in which the follower must have specified velocities during its stroke. 
If the follower is required only to move from one end of its stroke 
to the other in a given period of time, independently of all inter- 
mediate velocities, this form of construction may be readily applied. 
The principal difficulties to be met in the building of these cams, 
when the intermediate velocities are specified, are, first, the large 
time angles necessary for a desired follower motion, or, second, a 
comparatively large cam. The cause of these difficulties may be 
pointed out in Fig. 44, where it may be seen that the construction 
point, K h might have been so much further out radially that the 
perpendicular line, Ki K 2 , would have passed to the left of B and it 
would have been impossible to draw the smooth cam curve A G 
tangent successively to all the perpendiculars. The limiting prac- 
tical case appears when any three successive construction lines 
meet in a point, in which event the cam will have a sharp edge and 
be subject to excessive wear at that point. This subject is further 
considered in paragraph 106. 

82. If one is not limited in the time, or angle, in which the follower 
must do its work; or, if not limited in the size of the cam, this form 
of construction may be used for any set of velocity values so long 
as they produce a working surface which always curves outward 
or which has an edge which points outward. 

83. Exercise problem 7a. Required a radial periphery cam 
to operate an offset follower which has a flat surface perpen- 



50 ELEMENTARY CAMS 

dicular to the line of motion instead of a roller, the follower 
to move: 

(a) Up 3 units in 90° on the crank curve. 

(b) Down 3 " " 90° " " 

(c) Rest for 180°. 

Take cam disk to be one unit thick and the follower offset equal 
to two units measured from center of cam disk. Find and mark 
the locus of contact, also the size of the follower disk and the area 
of follower surface subject to wear. 

84. Cams for swinging follower arms. In the previous prob- 
lems the motion of the center of the follower roller has been in a 
straight line. When the center of the roller moves in a curve a 
different method of construction is used to advantage. Cams with 
swinging followers are illustrated in Figs. 45 and 46, the arc of 
swing A V of the follower having its extremities on a radial line in 
the former illustration; and on an arc which, continued, passes 
through the center of the cam in the latter illustration. These two 
forms of construction, although apparently differing in only a slight 
detail, give quite different results and each has its own particular 
field of usefulness. A comparison of the results will be given in 
paragraph 95 after a problem in each case has been worked out. 

85. Problem 8. Cam with swinging follower arm, roller 
contact — Extremities of swinging arc on radial line. Re- 
quired a radial periphery cam to operate a roller follower where the 
follower arm swings about a pivot and where the two extreme posi- 
tions of the center of the roller lie on a radial line. The chord of 
the swinging arc of the roller center is to be 4 units and the length 
of the follower arm 8 units. The follower arm to swing: 

(a) Out full distance in 90° on parabola curve. 

(b) In " " " 90° " crank 

(c) And to remain at rest for 180°. 

86. A different method of construction from any thus far em- 
ployed is used in problems of this kind because it gives the simplest 
and most accurate layout for the pitch surface. Briefly, the method 
to be used consists in revolving the follower through the 360° around 
the cam while the cam remains stationary, and drawing the follower 
in a number of its phases while on the way around. One of the 
phases is represented in full by the dash lines C w Y 2 F 3 in Fig. 45. 

87. The angle which causes pressure against the follower bear- 
ings is also different in this form of cam from any of the others. An 
inspection of Fig. 45 will show that, in general, the normal line of 



CAM PROBLEMS AND EXERCISE PROBLEMS 51 

pressure, AV at A, between the cam surface and the roller is not at 
right angles to the position of the follower arm, and, therefore, 
that the resultant total pressure has a component along the arm, 
tending to place it in compression and throwing a corresponding 
pressure on the follower bearing at C. The pressure angle at A is 
shown by —a, the minus sign indicating compression in the swinging 
arm. When K { is at K the pressure angle will be +c, the plus sign 
indicating tension in the follower arm. A disadvantage of the sign 
changing from + to — , etc., is that as soon as the bearings wear 
there will be noise at that point. 

88. The detail for the construction of problem 8 is taken up 
by computing the diameter of the pitch circle first, as in previous 
problems. This computation, however, serves only as a guide, for 
the assigned pressure angle will be both increased and decreased by 
amounts depending on the radius of the follower arm and the char- 
acteristics of the base curve which is used. For computing the 
pitch circle then, an assigned pressure angle factor for 30° will be 
assumed in the expectation that the final maximum angle will not 

exceed 40°. From formula 1, paragraph 29, d = 114.6 ■ — ^p — 

4 X 2 72 
= 17.62 for the parabola curve; and d = 114.6 — nTT" = 13.86 

for the crank curve assignment. The radius of the pitch circle is 
thus found to be 8.81 units. 

89. Having determined the radius D, Fig. 45, for the pitch circle, 
the given chord of 4 units is laid off with equal parts on each side 
of D, thus locating the ends of the swinging arc A V on the radial 
line D as required. With A and V as centers and a radius of 
8 units for the length of the follower arm, strike arcs which will 
intersect at C and give the fixed center for the follower arm. The 
arc A V, showing the path of the center of the follower roller is now 
drawn. 

90. Points on the pitch surface A Vi A 2 F are found, in brief, 
by revolving the arm C A around 0, swinging it a proper amount 
on its center C as it revolves. In detail this is accomplished by lay- 
ing off the arc C C 6 equal to 90°, and dividing it into a number of 
equal parts, say six. Divide the arc A J into three unequal parts, 
as at H and /, for the parabola curve. Lay off the points L and K 
in the same way. Then with C A as a radius and with d, C 2 . . . 
as centers draw the arcs passing through Hi, I\ . . . . Again, with 



52 ELEMENTARY CAMS 

as a center, swing arcs through H, I . . . until they meet the 
arcs already constructed. The intersections of these arcs, as at 
Hi, Iij Ji . . . will be the points on the desired pitch surface AV\. 
The determination of the pitch surface for the crank curve is found 
by laying off the second 90° assignment from C$ to C i2 and dividing 
it into six parts. The arc Ai Vi is divided by projecting the points 
U W ... of the crank circle to the points Ui Wi on the arc. The 
constructions for the points £/ 2 , W2 . . . are the same as for the 
previous part of the pitch surface, as described above. 

91. Exercise problem 8a. Required a radial periphery cam 
to operate a roller follower where the follower arm swings about a 
pivot and the two extremities of the swinging arc lie on a radial 
line. The 30° pressure angle factor to be used in computing the 
pitch circle radius. The chord of the swinging arc to be 3 units, 
the arm 9 units long, and to: 

(a) Swing out in }/$ revolution on the crank curve base. 

(b) Remain at rest for }/% revolution. 

(c) Swing in in }/% revolution on the parabola base. 

92. Problem 9. Cam with swinging follower arm, roller 
contact — Swinging arc, continued, passes through center of 
cam. Required a radial periphery cam to operate a roller follower 
where the follower arm swings about a pivot, and where the center 
of the follower roller moves on an arc which, continued, passes 
through the center of the cam. The chord of the swinging arc of 
the roller center is 4 units and the length of the follower arm 10 
units. The follower arm to swing: 

(a) Out full distance in 90° on parabola curve. 

(b) In " " " 90° " crank 

(c) To remain at rest for 180°. 

93. The procedure for this problem is the same as for Problem 8 
in all respects except the layout of the arc of swing for the center 
of the follower roller. The pitch circle is drawn with radius O J, 
Fig. 46. 

With the center of the cam O and the pitch point J as centers 
draw arcs which intersect at C, the radius being equal to the length 
of the follower arm. Lay off J A and J V equal to each other and 
so that a chord drawn from A to V equals the four units assigned. 
A bent rocker, A C E, is introduced in Fig. 46 simply to change the 
direction of motion. 

94. Exercise problem 9a. Required a radial periphery cam to 
operate a roller follower where the follower arm swings about a pivot, 



CAM PROBLEMS AND EXERCISE PROBLEMS 



53 



and where the center of the follower roller moves on an arc which, 
continued, passes through the center of rotation of the cam. Take 
the length of follower arm as 12 units and its angle of swing 30°. 
Required that the follower arm: 

(a) Swing out full distance in ^ revolution, on crank curve. 

(b) Remain at rest 34 revolution. 

(c) Swing in full distance in Y% revolution, on crank base. 




Fig. 45. — Problem 8, Cam with Swinging Follower Arm, Roller Contact — Extremi- 
ties of Swinging Arc on Radial Line 



95. Effect of location of swinging follower arm relatively 
to the cam. When the swinging follower arm is mounted so that 
the extremities of the arc of travel of roller center are on a radial 
line, as in Problem 8, the pressure angles on the out and in strokes 
will be approximately the same. When the follower roller center 



54 



ELEMENTARY CAMS 



moves on an arc which, continued, passes through the center of 
the cam, as in Problem 9, the pressure angle will be larger, on the 
average, on the one stroke than on the other. Consequently, the 
type shown in Problem 8 would have an advantage where equal 
amounts of work were to be done on both strokes, and the type 




Fig. 46.— Problem 9, Cam with Swinging Follower Arm, Roller Contact? — Swinging 
arc, Continued, Passes Through Center of Cam 



shown in Problem 9 would be of advantage where heavy work was 
to be done on one stroke only. Either the out or in stroke may 
be selected for heavy work, according to the position taken for the 
center C or G of the swinging arm, Fig. 46, the direction of turning 
of the cam being the same. In many cases the type shown in 
Problem 9 allows the pressure angle to be maintained on one of the 
strokes so that there is pressure in only one direction on the shaft C. 



CAM PROBLEMS AND EXERCISE PROBLEMS 55 

Cams operate smoother when "running off" than when "running 
on." A cam is said to be "running off" when the point of contact 
on the working surface of the cam is moving away from the fixed 
center of the swinging follower arm. A cam of the type illustrated 
in Problem 8 will have an axis of symmetry where the same data 
are assigned for the out and in stroke, whereas the cam illustrated 
in Problem 9 will be quite unsymmetrical for same data. 

96. Positive-drive face cams. The pitch surfaces for face 
cams are laid out in exactly the same manner as pitch surfaces for 
radial periphery cams. The only additional feature is that a working 
surface is drawn to touch each side of the roller. 

97. Problem 10. Face cam with swinging follower. Con- 
struct a face cam for a swinging follower arm, roller contact. Arm 
to be 12 units long and to swing through 30°. Required that the 
arm shall: 

(a) Swing full out on the combination curve while the cam makes 
% revolution. 

(b) Swing full in on the combination curve while the cam makes 
% revolution. 

98. In order to compute the radius of the pitch circle it is neces- 
sary to find the travel, or the approxi- 
mate travel, of the center of the follower 
roller. This is graphically done by mak- 
ing a separate sketch, as in Fig. 47, 
drawing the angle X Y Z equal to 30°, 
drawing the arc Z X with a radius of 12 
units, and measuring the chord Z X, 

Which is found tO be 6.2 Units. Or, this Fl «- 4 7-To Find Chord miai 

_ ' ure of Travel of Point on 

value may be found trigonometrically, swinging arm 

without any drawing, by taking 12 X 2 

sin 15° = 6.2. The radius of the pitch circle will then be: 

6.2 X 2.27 X y X 3-^ X | = 6.0 units. 

99. To construct the cam, the value just found is laid off at J, 
Fig. 48, and the pitch circle drawn. With the combination curve 
a cam chart, a partial one at least, must be drawn. To do this with 
least effort, select any point J' in line with the pitch point J and 
draw the line J' V at the given pressure angle, 30° in this case, 
until it is 6.2 units long. With V as a center, draw arc J' A' and 
also draw a tangent to it at J' and produce it to S/ where R S' equals 




56 



ELEMENTARY CAMS 



one-half A' V. The curve A' J' S' will be one-half of the desired 
base curve and will be sufficient to proceed with the construction of 
the cam. Divide the pitch line, 0-4, of the chart into four equal 
parts and draw verticals so locating H' } I', K f . . . . Project these 
points to H, I, K ... on the arc of travel of the center A of the 
roller. This construction will give practically a uniform swinging 




Fig. 48. — Problem 10, Face Cam with Swinging Follower 



velocity to the follower arm through twice the angle measured by 
the arc from J to S. Theoretically, the curve A' S' should be con- 
structed on the cylindrical surface A S instead of on its projected 
plane surface. It is, however, unnecessary to go into the detail 
of construction which this would involve because the difference in 
results between it and the more direct process explained above 
would be too small in practical cases to be measured by the thickness 
of the ordinary pencil line. 

With the points H, I, K . . . obtained as above, the remainder 
of the construction is the same in detail as described in connection 
with Problem 8. The reference letters are the same in both figures. 
The cam plate, in the face of which the groove for the roller is cut, 
is made circular in its boundary in order to give better balance and 
appearance. 



CAM PROBLEMS AND EXERCISE PROBLEMS 57 

100. Exercise problem 10a. Required a face cam for a swinging 

follower arm, roller contact. Arm to be 10 units long. Center of 
roller to swing through an arc whose chord is 4 units, and this arc, 
when continued, to pass through center of cam. The arm to: 

(a) Swing to the right on combination curve while cam turns 
180°. 

(b) Swing to the left on combination curve while cam turns 180°. 

101. Problem 11. Cam with swinging follower arm, sliding 
surface contact. Required a radial periphery cam to operate a 
swinging arm having a construction radius of 9 units. Sliding sur- 
face contact between cam and follower. The arm to: 

(a) Swing up 4 units on the crank curve base while the cam 
turns 120°. 

(b) Swing down 4 units on the crank curve base while the cam 
turns 120°. 

(c) Remain at rest for 120°. 

102. This type of cam and follower is illustrated in Fig. 49. The 
line of pressure between cam and follower is always normal to the 
follower surface and consequently there is no component of pressure 
in the bearing at C due to pressure angle. This cam is, therefore, 
independent of a pitch circle based on pressure angle, and the pitch 
circle may be taken any size. Where one has no special guide in 
assuming a starting size for the cam, the usual computation for 
pitch circle for a 30° pressure angle may give good average results. 
According to this, the pitch radius D will be, 

4 X 2.72 X 3 X 3-^ X \ = 5.2 units 

AV equals 4 units and A C equals 9 units. The point A is 
taken, for construction purposes, as a point on the follower arm 
where the angular velocity of the arm is measured. It will be 
at the points H, I, J ... on the arc A V at the end of equal suc- 
ceeding intervals of time. 

103. The method of constructing the cam in this problem is 
identical with the method used in Problem 8 in so far as the 
follower arm is swung around the cam, and its position with respect 
to the cam center at equal time intervals is drawn. The departure 
from the method of Problem 8 consists in drawing the cam outline 
as an envelope to these follower-arm positions. For example, in 
Fig. 49, at the end of the third time interval the pivot C has been 
revolved to C 3 and the point A of the follower arm has moved out 



58 



ELEMENTARY CAMS 



to J i. The point Ji is found at the intersection of two arcs, one 
obtained with Ci as a radius and C 3 as a center, and the other 
with J as a radius and as a center. 

When a number of positions of the follower arm, such as Cz J h 
have been obtained, the smoothest possible curve is drawn tangent 
successively to each of them, and this curve is the working surface 




Fig. 49. — Problem 11, Cam with Swinging Follower Arm, Sliding Contact 

of the cam. This curve is tangent to Cz J\ at M , and if the distance 
Cz M is laid off at C M h the point Mi will be the actual point of 
tangency between the cam surface and follower arm when the arm 
is halfway through its swing, or when A is at J. Similarly when 
Cq is at C the point of tangency between cam and follower arm will 
be at Ni. 

104. The locus of the point of contact between the cam and 
follower, relatively to the frame of the machine, is shown by the 



CAM PROBLEMS AND EXERCISE PROBLEMS 



59 



dash closed curve through M\ and Ni. By drawing arcs tangent 
at the extremities of this dash curve, using C as a center in both 
cases, the points F and G on the follower surface are obtained and 
the distance F G will be the part of the follower exposed to wear 
from the rubbing of the cam. This part of the follower arm may be 
designed with a shoe, as indicated, which may be replaced when 
worn. 

105. It should be specially noted that the shortest radius of the 
cam is not A, but B. The point B is found by drawing a per- 
pendicular to C G through 0. 

The very decided lack of symmetry should also be noted > the 
curve B L being used to lift the arm, and the curve L E to lower 
the arm, the swinging velocities of the arm being the same in both 
directions. 

106. Data limited for followers with sliding surface con- 
tact. The data for this type of cam construction are extremely 
limited when the swinging velocity of the arm is assigned. The 
limitations are that the working surface of the cam must be drawn 
tangent to every construction line in succession, and that it must be 
convex externally at all points. In most arbitrary assignments of 
data the construction line through Cg, for example, would intersect 
the line through d before it cut the line through Cg. In this case 
it would be impossible to draw a smooth working curve tangent, 
successively, to the lines through d, C%, and Cg. This is illustrated 
more clearly in Fig. 51 and will be more evident after the limiting- 
case is described. 

The limiting case for flat surface followers with sliding contact 
occurs where three or more of the construction lines meet in a point, 
as at N in Fig. 50. In this case the working surface of the cam 




Fig. 50. — Limiting Case for Straight Edge Fig. 51. — Impossible Case for Straight 
Follower with Sliding Contact Edge Follower with Sliding Contact 



60 



ELEMENTARY CAMS 



would have a sharp edge. In this type of cam it is necessary to 
use more construction lines than in other types, because it is pos- 
sible to have the construction lines so far apart that such a case as 
is shown in Fig. 51 might not evidence itself at all. For example, 
if the distance C 9 C7 were the unit space for construction lines, in- 
stead of C 9 C 8 , the smooth convex curve F N L could be drawn 
tangent to lines through C 9 , C 7 . . . without the error showing 
itself. 

107. If it is required of this cam only that it shall swing a follower 
arm through a given angle in a given time, without regard to the 




Fig. 50. — (Duplicate.) Limiting Case for 
Straight Edge Follower with Sliding 
Contact 



Fig. 51. — (Duplicate.) Impossible Case 
for Straight Edge Follower with 
Sliding Contact 



intermediate velocities of the arm, it may be as widely used as any 
other type of cam. In this case only the innermost and outermost 
positions of the arm would be drawn, as at C A, C& V h and C i2 E, 
Fig. 49, and a smooth convex curve drawn tangent to these lines. 
Such construction, however, might give an irregular or jerky motion 
to the follower. Whether it did or not could be readily determined 
by laying off a number of equal divisions, as at C h C 2 . . . C i2 ; 
drawing lines, such as C% Ji, tangent to the assumed smooth convex 
working surface; and revolving C 3 Ji back to C J. After doing 
this with other construction lines a series of points, such as H, I, J 
. . . would be determined and the spaces between them would 
represent the distances traveled by A on the follower arm during 
successive equal intervals of time. 

108. Exercise problem 11a. Required a radial periphery cam 
for a swinging follower arm, sliding surface contact. Arm to be 
10 units long to the point which is used to measure the angular 
velocity, and this point to move through an arc which is measured 
by a chord of 4 units. The arm is to : 



CAM PROBLEMS AND EXERCISE PROBLEMS Gl 

(a) Swing full out with uniform acceleration and retardation 
while the cam turns % revolution. 

(b) Swing in with the same angular motion in Y% revolution. 

(c) Remain stationary for J^ revolution of the cam. 

109. Toe and w t iper cams. In this form of cam construction 
the cam or " wiper" OC, Fig. 52, oscillates or swings back and 
forth through an angle of 120° or less, instead of rotating con- 
tinuously the full 360° as it does in all cams thus far considered. 
The follower or "toe" A W is usually a narrow flat strip resting on 
the curved periphery of the cam, and moving straight up and down. 
There is sliding action between the wiper and the toe. 

110. Problem 12. Toe and wiper cam. Required a wiper cam 
to operate a flat toe follower which shall move: 

(a) Up 4 units with uniform acceleration all the way while the 
cam turns counterclockwise 45° with uniform angular velocity. 

(b) Down 4 units with uniform retardation all the way while 
the cam turns clockwise 45° with uniform angular velocity. 

111. The detail of construction for this class of problem is iden- 
tical with that described for the mushroom cam in Problem 7, 
it being observed that the two cams differ only in that the mush- 
room cam turns through the full 360° instead of 45° as in this problem, 
and the mushroom follower is circular instead of rectangular. Neither 
of these differences nor the offset of the mushroom follower affect 
the similarity of construction for the two types of cams. There- 
fore, only a brief review of the general method of construction for 
the present problem will be given here. 

112. Inasmuch as the line of pressure between cam and follower 
is always parallel to the direction of motion of the follower in prob- 
lems such as this, there is no pressure angle in the ordinary sense. 
If a computation for size of cam is made in the usual way, the radius 
of the pitch circle will figure to be unnecessarily large, due princi- 
pally to the fact that only a 45° degree turn of the cam is allowed 
for the upward motion of the follower. 

A radius O A, Fig. 52, which allows for radius of shaft, thickness 
of hub, etc., is assumed, and the follower motion of 4 units is laid 
off at A V. This distance is divided into four unequal parts at 
H, I . . . which are to each other as 1, 3, 5, and 7, thus giving 
uniform acceleration all the way up. The angle A B of the cam 
is laid off 45° and is divided into four equal time parts. The follower 
or toe surface A W is then moved up the distance A H and revolved 
through the angle A 1 to the position H x H 2 which is marked. Simi- 



62 



ELEMENTARY CAMS 



larly A W is next moved to I I 3 and revolved to h 7 2 . The smooth- 
est possible convex curve is then drawn to the lines Hi H 2 , I\ h . . . 
and this curve becomes the working surface of the wiper. 

The necessary working length for the wiper is found to be A V 2 , 
and, adding a small arbitrary distance, V 2 C, the total length is taken 




Fig. 52. — Problem 12, Toe and Wiper Cam 

as A C. The total length of the toe A W will be equal to V\ C. 
The long dash lines in Fig. 52 indicate the highest position of the 
toe and wiper, and the short dash-line curve marks the locus of 
contact between the wiper and toe. This curve is obtained by 
making, for example, J J 3 equal to Ji J 2 - 

113. Modifications of the toe and wiper cam. The toe and 
wiper cam constructions are commonly used. In the present ele- 
mentary problems the cam or wiper is assumed to oscillate with 
uniform angular velocity, whereas in practice it usually has a variable 
angular velocity due to the fact that it is operated through a rod 
which is connected at the driving end to a crank pin or eccentric 



CAM PROBLEMS AND EXERCISE PROBLEMS 03 

whose diameter of action corresponds to the swing of the wiper 
cam. The follower toe may be built with a curved instead of a 
straight line, by a slight modification in detail which consists in draw- 
ing the curved toe line in place of the straight lines, Hi 11*, l x I 2 . . . 
as shown in Fig. 52. These points, together with a consideration of 
the amount of slip between the surfaces in this type of cam and a 
discussion of the necessary modification to secure pure rolling in 
cams of this general appearance, are subjects for more advanced 
work than is covered by the present elementary problems. 

114. Exercise problem 12a. Required a wiper cam to operate 
a flat toe follower which shall move: 

(a) Up 3 units with uniform velocity while the cam turns G0° 
in a counterclockwise direction with uniform angular velocity. 

(b) Down 3 units with uniform velocity while the cam turns G0° 
in a clockwise direction with uniform angular velocity. 

115. Yoke cams. Yoke cams are simple radial periphery cams 
in which two points of the follower, instead of one, are in contact 
with the working surface. The contact points are usually diametri- 
cally opposite to each other. Roller contact is generally used and 
the centers of the rollers are a fixed distance apart. The yoke cam 
gives positive motion in both directions, and does not depend on a 
spring or on gravity to return the follower as do all other cams 
thus far considered, excepting the face cam. 

116. Problem 13. Single-disk yoke cam. Required a single 
radial cam to operate a yoke follower with a maximum pressure 
angle of 30° : 

(a) Out 4 units in 45° turn of the cam, on crank curve. 

(b) In 4 " " 90 c 

(c) Out 4 " " 45 c 

117. With a single radial cam for a yoke follower, data may be 
assigned only within the first 180°. The reason for this will appear 
presently. 

Compute the radius of pitch circle as in ordinary radial cam 
problems. It is found to be 13.86 units and is laid off at D, Fig. 53. 
The pitch surface, A D1V1A1 V 2 , is found in the usual way. Then 
the diametral distance, AV 2 , will be the fixed distance between the 
centers of the rollers, and if this distance is laid off on diametral lines, 
as from h, Ki . . ., the points W, X ... on the complementary 
pitch surface will be located. A size of roller A B is next assumed 
and the working surface B B 2 is constructed. The maximum radius 
of the working surface is finally located, as at B 2 . A small amount 



jO u 
£0 tt 



64 



ELEMENTARY CAMS 



is added to this for clearance and the total laid off at Z, thus giving 
the width of yoke necessary for an enclosed cam. 

118. Limited application of single-disk yoke cam. In yoke 
cams constructed from a single disk the data are limited in two ways : 

First, data can be assigned for the first 180° only, because the 
pitch surface for the second 180° must be complementary to the 
pitch surface in the first 180°. 

Secondly, the complementary pitch surface cannot approach any 
nearer to the center of rotation of the cam than does the pitch surface 



TT-W4 1 - — 






np 



±: rrh — 

" T H" ? 




Fig. 53.— Problem 13, Single-Disk Yoke Cam 



in the first 180°, otherwise the follower will have a greater motion 
than that which was assigned to it. 

To illustrate this second case, assume that item (c) had been 
changed in the data for Problem 13 so as to specify that the follower 
should remain at rest while the cam turns 45°. Then the pitch 
surface of the cam for the first 180° would have been AViAiC, 
Fig. 54, instead of AVi A x V 2 . The diametral distance A C would 
then have been the distance between roller centers, and would have 
been also the distance used in determining the complementary 
pitch surface C E x A s A which, it will be noted, approaches closer 
to O than does A Vi C. When E x of the complementary surface 



CAM PROBLEMS AND EXERCISE PROBLEMS 



65 




reaches the center line D, the center A of the roller will be at E 
and the roller will have traveled the distance A E in addition to the 
travel A V which was assigned. Furthermore, the pressure angle 
will be very high when F crosses the line D 2 . With the data 
which gives the pitch surface AV\C, the yoke follower will move 
just twice the assigned distance. This double motion will not be 
continuous, as the follower 
will be at rest for a definite 
period represented by A X C. 
Even if the data were such 
that A i should fall at C there 
would be a momentary pe- 
riod of rest for the follower 
at the middle of its stroke. 

Summing up, the desired 
travel, pressure angle, and 
follower velocity will be ob- 
tained in single-disk yoke 
cams, only when the data 
are such as to have the fol- 
lower at the extreme oppo- 
site ends of its stroke at the 

zero and 180° phases. In other cases increased travel, increased 
pressure angle, and irregular follower velocities will have to be 
considered. 

All of the limitations of the single-disk yoke cam may be avoided 
by using the double disk cam as illustrated in Problem 14. 

119. Exercise problem 13a. Required a single-disk radial cam 
to operate a yoke follower with a maximum pressure angle of 30°: 

(a) In 6 units in 60° turn of the cam on parabola curve. 

(b) Out 6 " " 45° " " " " " 

(c) At rest for 30° " " " " 

(d) In 6 units in 45° " " " " " 

120. Problem 14. Double-disk yoke cam. Required a double- 
disk cam to operate a yoke follower with a maximum pressure angle 
of 30°: 

(a) To the right 6 units in 150° turn of the cam, on the crank 
curve. 

(b) To the left 6 units in 90° turn of the cam, on the crank curve. 

(c) To remain stationary for 120° turn of the cam. 



Fig. 54.- 



i illustrating limited application of 
Single-Disk Yoke Cam 



66 



ELEMENTARY CAMS 



121. The detail of construction for the primary disk is the same 
as in previous problems involving radial cams. In this problem, then, 

the radius of the pitch circle is 6 X 2.72 X 4 X o"T7 X ~k = 10.4 

units and this is laid off at D, Fig. 55. The forward driving pitch 
surface, A Hi Vi 7i A , is constructed in the regular way as indicated 
by the construction lines. 

122. The diameter D C of the pitch circle is next taken as a 
constant and its length is laid off on diametral lines from successive 




Fig. 55. — Problem 14, Double-Disk Yoke Cams, Detail Construction 



points on the primary pitch surface, thus giving the secondary or 
return pitch surface. For example, the point P on the secondary 
surface is found by making A P = DC; the point M by making 
Hi M = D C. . . . This second cam disk has a pressure angle of 
30° at D A , the same as the primary disk has at D 2 . Had any diam- 
etral length other than D C been taken in this problem as a constant 
for constructing the second cam, the pressure angle at D 4 would 
have been greater or less than the assigned 30°. It does "not follow 
that the diameter of the pitch circle should be used as a constant 
for generating the complementary cam. The determining factor, 
in selecting a constant diametral length is that the maximum pres- 
sure angle on the second cam should not exceed the assigned value. 



CAM PROBLEMS AND EXERCISE PROBLEMS 



67 



123. To avoid intricate line work, only the detail drawing for 
the construction of the pitch surfaces for this problem is shown 
in Fig. 55. The pitch surfaces are then redrawn in Fig. 56 and 
the working surfaces and the yoke constructed. 

The working surface of the primary or forward-driving cam is 
shown at B E F G B, Fig. 56, and is constructed in the same way as 




Fig. 56. — Problem 14, Double-Disk Yoke Cams Showing Strap Yoke and Rollers 



in previous problems by drawing it as an envelope to successive 
roller positions. The working surface of the return cam is shown 
at S Q R S. A special caution to be observed at this point is that 
the working surface of the second cam cannot be obtained directly 
from the working surface of the first cam by using the diametral 
constant; the second cam pitch surface must be obtained first. 

124. The form of yoke in yoke cams may vary, as illustrated for 
example by the box type which encloses the cam, Fig. 53, and by 
the strap type, Fig. 56. In the latter illustration the strap W X 
has a longitudinal slot T U permitting it to move back and forth 
astride the shaft without interference. The guide arms of the 



68 ELEMENTARY CAMS 

yoke are shown at Y and Z. In all yoke constructions it is desirable 
to have all the forces acting in as nearly a straight line, or in a plane, 
as possible.' In Fig. 53 this is obtained, as may be noted in the 
top view where the longitudinal center lines of cam disk, cam roller, 
yoke and yoke guides are all in the same plane. In Fig. 56 the yoke 
guides, Y r and Z r , are placed in a line lying between the cam disks, 
B f and S f , so as to have the forces balanced to a greater degree than 
they would be if the guides were in line with the strap W f X' '. 

125. Exercise problem 14a. Required a double-disk cam to 
operate a yoke follower with a maximum pressure angle of 30°, as 
follows : 

(a) To the right 4 units in 90° on the parabola base. 

(b) Dwell for 30°. 

(c) To the right 4 " " 105° " " 

(d) " " left 8 " " 135° " " 

126 Problem 15. Cylindrical cam with follow t er that 
moves in a straight line. Required a cylindrical cam to operate 
a reciprocating follower rod : 

(a) To the right 4 units in 120° on the crank curve. 

(b) " " left 4 " " 120° " " 

(c) " dwell 120°. 

The maximum surface pressure angle to be 30°. 

127. The size of cylinder is found by a computation similar to 
that for radial cams, and in this problem the radius of the cylinder is, 

4 X 2.72 X 3 X t^tt X 2 = 5.2 units. 

This distance is laid off at O f A r in Fig. 57, and the circle drawn. 
The distance A V, the travel of the follower, is laid off equal to 
4 units and subdivided, according to the crank circle, at H, I . . . 
The radius of the follower pin is assumed as at A S and this distance 
is laid off at S C, thus locating the edge of the cylinder. Make 
V D equal to A C. The circle representing the cylinder is next 
divided into three 120° divisions at A f , M f , and Q', as specified. 
A r M r is divided into six equal parts by the points H f , V . . . 
which are projected over to meet the vertical construction lines 
through H, I . . .at H 2 , li. . . . The latter points mark a curve on 
the surface of the cylinder. This curve is a guide for the center of 
the tool which cuts the groove. The finding of this curve and the 
construction of the follower pin and rod constitute the remaining 
essential work on this problem. If it is desired to show the groove 



CAM PROBLEMS AND EXERCISE PROBLEMS 



69 



itself, the directions in paragraph 134 will give an approximate 
method. The follower pin is attached to a follower rod X which 
is guided by the bearings Y and Z. The assigned pressure angle 
of 30° is shown in its true size at D J G; J D being parallel to the 
direction of motion of the follower rod, and J G being a normal 
to the cutting-tool curve M N J P. . . . In general, the pressure 
angle will not show in its true size, and if it is then desired to illus- 
trate it, the cylinder may, in effect, be revolved until the correct 
point of the cutting-tool curve is projected on the horizontal center 
line. The exact point E where the cutting-tool curve comes tangent 
to the bottom line of the cylinder may be found by locating E\ 
relatively to Ki and Li, the same as E' is located relatively to K r 
and Z/, and projecting E x down to E. 

A small clearance is allowed between the end B f of the pin and the 
inner surface of the groove, which is represented by the dash circle 
passing through F r . 

128. Refinements in cylindrical cam design. It will be 
noted that the " maximum surface pressure angle" was given in the 
data for this problem instead of the term " maximum pressure 
angle" that has been used thus far. The reason for this is that 
the pressure angle varies along the length of the pin and is always 
greatest at the outer end, that is, at the point B in Fig. 57. This is 
not important in most practical cases. Further, the term "pitch 
cylinder" is not mentioned in the simple form of practical construc- 




E 'K' M£' L' 

Fig. 57. — Problem 15, Cylindrical Cam with Follower Sliding in a Straight Line 



70 ELEMENTARY CAMS 

tion here used. Since the pitch cylinder should pass through the 
point where maximum pressure angle exists, the pitch cylinder 
in cams of this type would be one having a radius O f B' '. The pitch 
surface of the cylindrical cam would be a warped surface, known 
as the right helicoid, and the intersection of this surface with the 
surface of the cylinder is the curve R A E P R and is the guide 
curve for the cutting tool in milling out the groove for the pin. 
The sides of the groove are the working surfaces of the cam; they 
are indicated in the sectioned part of the front view of Fig. 57. 

More exact methods for drawing the sides of the groove in a 
cylindrical cam, together with a more exact method for determining 
the maximum pressure angle, involve a knowledge of projections 
and an intricacy in drawing that make such work a proper subject 
for advanced study, and it will therefore be omitted in this elemen- 
tary treatment, as it is totally unnecessary in most practical work. 

129. Exercise problem 15a. Required a cylindrical cam to 
operate a sliding follower rod, with a maximum pressure angle of 40°: 

(a) 6 units to the right in 90° on the parabola base. 

(b) 6 " " " left " 270° " " 

130. Problem 16. Cylindrical cam w t ith swinging follower. 
Required a cylindrical cam to operate a swinging follower arm: 

(a) To the left 40° in 120° turn of the cam on the crank curve. 

(b) " " right 40° " 120° " " " " " " 

(c) Dwell for 120° turn of the cam. 

The length of the follower arm is to be 9 units and the approximate 
maximum pressure angle is to be 30°. 

131. The diameter for the cylindrical surface is found in the 
same manner as the diameter of the pitch circle in radial periphery 
cams. The data in this problem do not give directly the travel of 

the follower and so this value must be 
found first. The chord of a 40° arc hav- 
ing 9 units radius will be 

9 X 2 X sin 20° = 9 X 2 X .342 = 6.2 units. 

\D L_I ^< | 

G If a trigonometrical table is not at 

Fig. 58. — Determining Fol- , , ,, , , . . 

lower travel in swinging hand the arc may be drawn out as m 

dSSe F ter C ° MPUTING PlTCH Fi S- 58 where half the S iven an ^ le is laid 

out at AY J by the simple expedient of 

subdividing a 30° arc by means of a dividers. The half chord 
A D is drawn and measured. It is equal to 3.1 units, thus mak- 
ing the chord of the whole arc of travel equal to 6.2 units. 




CAM PROBLEMS AND EXERCISE PROBLEMS 



71 



This value is used in obtaining the diameter of the surface of the cyl- 
inder as follows: 

G.2 X 2.72 X 3 X 



3.14 



16.12 units. 



The circle A' Q r M f , Fig. 59, is drawn with a radius of 8.06 units. 

132. The 120° angles assigned in the data are next laid out but 
not from the center line R r as in previous problems. In mechan- 
isms of all kinds where there is a Swinging follower, it is a rule, unless 
otherwise specified, that the swinging pin should be the same dis- 




Fig. 59. — Problem 16, Cylindrical Cam with Swinging Follower Arm 



tance above a center line at the middle of its swing as it is below 
at the two extremities of its swing. In this case, then, the point G, 
Fig. 58, will be marked midway between J and D and the distance 
G J laid off at G J in Fig. 59. Y will be the center of swing of the 
follower arm and the arc of swing of the follower pin will be A J V. 
J will be as much above the center line as A and V are below. The 
practical advantage of this detail in the layout is that it gives a 
maximum bearing length between the follower pin and the side of 
the groove. 

133. The arc A J V, Fig. 59, is next divided at the points marked 
H, I . . . according to the crank curve assignment, and vertical 
construction lines are drawn through these points. 

The point A is now projected to A' and the radial line, A' 0, is 
drawn. This becomes the base line from which to lay off the three 



72 ELEMENTARY CAMS 

assigned timing angles of 120°, as shown at A' M', M' Q', and 
Q f A'. The arc A' M' is next divided into the desired number 
of equal construction parts, as at i7 3 , 7 3 , J3. . . . 

When Hz reaches A', the pin A will have swung not only over 
to H } but it will have moved up the distance A' H' measured on the 
surface of the cylinder. Therefore, when H z reaches A', it is the 
line through H$ (i7 3 H$ = A' H') on the groove center line that will 
be in contact with the pin center line. For this reason H b , instead 
of i7 3 , is projected over to meet the construction line at H 2 . This 
latter point is on the guide curve for the cutting tool on the surface 
of the cylinder. Other points are found in the same way. Time 
may be saved by marking the points A' H' I' J' on the straight edge 
of a piece of paper and transferring these marks at one time so as 
to obtain the points 7 5 , J 5 . . . P 5 . . . . 

134. If it is required to show the surface bounding lines of the 
side of the groove it may be done quickly, although approximately, 
by laying off on a horizontal line, as at 7 2 , the points 7 4 and 7 6 at 
distances equal to the radius of the pin. These will represent points 
on the curve. If it is required to show the bottom lines of the 
groove it may be done by projecting from 7 7 and finding, for 
example, the point 7 8 in the same way as 7 4 was found. 

135. Exercise problem 16a. Required a cylindrical cam to 
operate a swinging follower arm: 

(a) To the right 6 units (measured on chord of follower pin arc) 
while cam turns 150°. 

(b) Dwell while cam turns 120°. 

(c) To the left 6 units while cam turns 90°. 

The follower arm to be 8 units long and its rate of swinging to be 
controlled by the crank curve with a maximum approximate pres- 
sure angle of 40°. 

136. Chart method for laying out a cylindrical cam with 
a swinging follower arm. This method is illustrated in Figs. 60 
and 61. The data in this problem will be taken the same as in 
Problem 16, namely, that a follower arm of 9 units length shall: 
Swing through an angle of 40° to the left while the cam turns 120°; 
through the same angle to the right while the cam turns 120°, on 
the crank curve in both directions; remain stationary while the 
cam turns 120°. The maximum pressure angle is to be approxi- 
mately 30°. 

137. To find the length of the chart, the chord that measures 
the arc of swing of the follower pin is first determined to be 6.2 



CAM PROBLEMS AND EXERCISE PROBLEMS 



73 



K:/ 



M< 



y* 



K 



Qi 



units as explained in paragraph L31. The length 

of chart is 

6.2 X 2.72 X3 = 50.G units, 

and this is laid off at J J h Fig. 60. The length 
of the follower arm is then laid off at J Y, and 
the follower-pin arc A V drawn. This arc is 
subdivided at H, I . . . according to the crank 
curve. The distance Y F 6 is then laid off to 
represent 120° and its length will be equal to 
one-third the length of the chart. As many 
construction points as were used from A to V 
are then laid off between Y and Y G . With 
these as centers and YA as a radius draw a series 
of arcs to which the points H, I . . . are pro- 
jected, thus giving the base curve through the 
points H h 1 1. . . . Tangent to the series of 
arcs on the chart draw straight lines and mark 
the intercepts H4H2, h h- • . . 

138. Upon completing the chart, the surface 
of the cam is drawn as in Fig. 61, with a diam- 



eter E' T' = |^j = 16.12. 



The width C N of 



the cylinder may be taken equal to the chord 
A V of the arc of swing of the follower pin, plus 
twice the diameter of the pin. 

139. The simplest general plan for trans- 
ferring the cam chart to the surface of the 
cam is to consider the chart lines to be on a 
strip of paper, and that this paper is simply 
wound around the cylindrical surface of the 
cam, starting the point G of the chart at G on 
the center line of the cam. G on the chart is 
midway between J and D. Then the points 
H 2 , li . . . of the base curve in Fig. 60 will fall 
at H 2 , 1 2, in Fig. 61, giving the surface guide 
curve for the center of the cutting tool. 

140. The detail necessary to actually lo- 
cate the points H 2 , 1 2 in Fig. 61 is accomplished 
by projecting J to J' and laying off the as- 



74 



ELEMENTARY CAMS 



signed 120° divisions, and also the subdivisions from this latter 
point. The 120° divisions are shown at M', Q', J'; the equal 
subdivisions at H 3 1 3 . . . . Prom these latter points, lines are pro- 
jected to the front view and the lengths H± H 2 , 1 4 1 2 are transferred 




Fig. 61.— Cylindrical Cam with Swinging Follower Drawn from Chart 



from Fig. 60. To find the point of tangency at E, make i£ 4 E\ 
of Fig. 60 equal to K 3 E' of Fig. 61, then draw E l E in Fig. 60 
and lay off this distance from the center line G Y in Fig. 61, thus 
giving the point E. To find the point of tangency at M, lay off 
at M ' M s a distance equal to the chart distance from M 2 to M 
in Fig. 60 and project M z of Fig. 61 to M. 



SECTION IV.— TIMING AND INTERFERENCE OF CAMS 



a 



141. In machines where two or more cams are emploj'ed it is 
generally necessary to lay down a preliminary diagram showing 
the relative times of starting and stopping of the several cams, in 
order to be assured that the various operations will take place in 
proper sequence and at proper intervals. The same preliminary 
diagram is also used to avoid interference and to make clearance 
allowances for follower rods whose paths cross each other. 

142. Problem 17. Cam timing and interference. Required 
two cams that will operate the follower rods A and E, Fig. 62, lying 
in the same plane, so that: 

(a) Rod A shall move 16 units to D, dwell for 30°, return 8 units 
to B and again dwell 30°, all to 
take place in 180° turn of the 
cam. The cam to produce the 
same motions in the second 180° 
but in reverse order. 

(b) Rod E shall cross path 
of rod A and move 4 units be- 
yond it and back again during 
the time that rod A is moving 
from D to B to D. 

All motions to be on the 
crank curve with maximum pres- 
sure angles of 40°. 

143. Before taking up the 
solution of this problem in de- 
tail it should be noted: 1st, that fig. G2.— problem 17, preliminary layout 
any convenient type of cam may £££ FOR PaOBLEM IN CAM lNTER " 
be used in problems of this kind ; 

2d, that usually only general motions of followers or objects are given 
in the preliminary data, as above, and that the cam designer must 
supply data and restate the problem in terms of angles for each of 
the movements after studying the preliminary data with the aid of 
a timing diagram. 

144. The first step leading to a restatement of the problem is to 
determine the number of degrees in which rod A may move the 

75 



76 



ELEMENTARY CAMS 



16 units, and also the number of degrees in which it may move 

the 8 units in order that the pressure angle will be 40° in both cases. 

Since there are two 30° dwells in the first 180° there will be 120° 

left for the two motions of which the first 

-j n 

motion will require ^r °f 120° or 80°, and the 

second, 40°. The length of chart for cam A 

may now be computed as 16 X 1.87 X t^t 

= 134.6 and laid off as at A A h Fig. 63. The 
height of the chart AD is 16 units. The 
chart is next divided into degrees of any con- 
venient unit, 0, 10°, 20° . . . being used in 
this case. For the present the base line may 
be made up of a series of straight lines as at A 
D h D, D 2 , D 2 Bl . . . 

145. The amount of clearance between 
the moving arms must now be decided upon. 
Let it be the designer's judgment that the end 
of the follower rod E should lie at rest 1 unit 
to the left of rod A as shown in Fig. 64, and 
that rod E should not begin to move until 
the rod A is one unit out of the way. Then 
A will be at O , Fig. 64, moving down, when 
E starts, assuming the rod E to be 3 units 
wide and that it is so placed that its top edge 
is one unit below D. The point C is then 5 
units from the top of the stroke and if this 
distance is laid off in Fig. 63, as shown, the 
line C Ci is obtained cutting the crank curve, 
which should now be drawn at C. C is at 
the 133° point and this, then, is the time when 
the follower E should start to move. 

146. The total motion for rod E is 4 + 5 + 
1 = 10 units, assuming width of rod A to be 5 
units. The time during which this motion can 
take place, outward, is 180° - 133° = 47° as 

represented at E x E 2 , Fig. 63. If the crank curve E\ F is now drawn it 
will be intersected by the one-unit clearance line GiGatG which rep- 
resents, in this case, a rotation of approximately 11° of the cam that 
drives rod E. The total clearance for the two rods which cross each 




Fig. 63.— Problem 17, 
Timing Diagram for 
Avoiding Interfer- 
ence of Cams 



TIMING AND INTERFERENCE OF CAMS 



77 



other's paths is now found to be 3° for cam follower A and 11° for cam 
follower E, or 14° of the machine cycle. These clearances are in- 
dicated in Fig. 63. If it is the judgmenl of the designer thai errors 
in cutting keyways and in assembling, and thai the wear of the parts 
will fall within these limits, the cams may now be drawn. 

147. The cam chart for cam E was made the same length as 



rarv 

i;j„L_. J. i 

u T 1 




Fig. 64. — Problem 17, Design for Definite Timing and Non-Interference of 
Cams Operating in Same Plane 



the chart for cam A in order to make clearance allowance. The 
true length of this chart, for a 40° pressure angle, would be: 

360 



10 X 1.87 X 



47 



143.2 units, 



instead of 134.6 as now drawn. If an exact clearance allowance 

in degrees were required, it would be necessary to redraw the crank 

47 
curve Ei F, making the distance EiE 2 equal to ^x of 143.2. It is now 

47 

o«7j of 134.6. With a new and exact drawing the crank curve E\ G 



78 ELEMENTARY CAMS 

would not rise quite so rapidly and the intercept at G would show a 
small fraction over the 11° taken above. In some problems where the 
lengths of the true charts differ considerably it may be necessary to 
redraw this part of the base curve to be sufficiently accurate in 
obtaining the clearance in degrees. 

148. The radius of the pitch circle for the cam operating rod A 

will be „ q = 21.4 units as drawn at H I, Fig. 64. The pitch 

surface of the cam and the working surface are drawn in the same 
way as the ordinary radial cams in previous problems. The length 
of the rod A± A may be assumed. 

The radius for the pitch circle for the cam operating rod E will be 

143.2 

n ' = 22.8 and this is laid off at M S. The location of M and the 
6.28 

length of the rod N E will either enter into the layout of the frame- 
work of the machine in a practical problem, or will be determined 

by the framework if pre- 



«- ? viously laid out. In the 

D ci present case it will only 

!_ I |__ be necessary, in deter- 

- 4 >] ! \t 4 — J 1 1 4 — >| I mining the length of rod 

— I — t- N E and the position 



i ii i 

B C of M, to make certain 

i Ii i 
i j i 1. 



that the shafts M and H 

Fig. 65.-Probi.em 17a, Diagram Showing Appli- are Sufficiently far apart 

cation of data to keep the cams from 

striking when turning. 

149. Location of keyw t ays. It is important to locate the 
keyway exactly by giving its position in degrees so as not to destroy 
the clearance values already made. Since the working surfaces of 
cams frequently approach close to the hub or shaft it is a good 
plan to place the keyway at the center of the longest lobe of the 
cam, as illustrated in both cams. 

150. Exercise problem 17a. Assume a stack of blocks at A, 
Fig. 65. Required that the bottom block shall be delivered with 
one stroke at C, the next block at D, being moved first to B and 
then to Z>, the next block at C, the next at D, etc. Let the sizes of 
the blocks and the distances they must be moved be as shown in 
Fig. 65. Lay out cam mechanism to secure this result, keeping 
the maximum pressure angle at 30°. 



SECTION V.— CAMS FOR REPRODUCING GIVEN CURVES 

OR FIGURES 

151. Problem 18. Cam mechanism for drawing an ellipse. 
Required a cam mechanism that will reproduce the ellipse A C B D 
in Fig. 66, the marking point to move slowly at the extremities A 
and B of the major axis and rapidly at C and Z>, the rate of increase 
and decrease of velocity being uniform. 

152. Divide A C into three parts which are to each other as 1, 
3, and 5; C B into three parts which are as 5, 3, and 1 ... in order 
that the marking point shall move through increasing spaces in 
equal times in moving from A to C. . . . For greater accuracy A C 
would be divided into a greater number of parts. 

153. In devising the mechanism assume that the marking point 
shall be at the end of a rod which shall be controlled by two com- 
ponent motions that are horizontal and vertical, or nearly so. This 
suggests the rod A E F, with marking point at A, with horizontal 
motion supplied from a bent rocker attached at F and with vertical 
motion supplied from a reversing straight arm rocker L K J, attached 
through a link E J at the point E. The lengths of the links and of 
the arms of the rockers, and the positions of the fixed centers of the 
rockers will have to be assumed, the lengths of the arms and links 
being such that none of them will have to swing through more than 
60°. With more than 60° swing the angle between an arm and a link 
is liable to become too acute for smooth running. Where rocker 
arms are connected to links the ends of the rocker should, in general, 
swing equal distances above and below the center line of the link's 
motion, as for example, the points F and 6 on the arc of swing of F 
should be as much above the line A M as the point 3 is below. Also 
the arc 3 J 9 should swing equally on each side of J T in order to 
secure best average pressure angles for the mechanism. 

154. Let each of the rocker arms be assumed to be controlled by 
single-acting radial cams. The center of roller H will be required 
to swing on an arc 6 H which, continued, passes through M. This 
gives small pressure angles while A is traveling to B, especially when 
A is at C and is moving fastest. It gives large pressure angle, how- 
ever, while A is traveling from B to D to A. If A is assumed to do 
heavy work along AC B and to run light along B D A this is the 

79 



80 ELEMENTARY CAMS 

better arrangement. If A did the same work on both strokes it 
would be better to place the rocker arm G H so that H and 6 rested 
on a radial line. The center of roller L will be assumed to travel 
on an arc whose extremities are on a radial line, or nearly so. 

With A F as a radius and A, 1, 2 . . . as centers, strike short 
arcs intersecting F 6 at F, 1 , 2 . . . numbering the arcs as soon as 
drawn to avoid confusion later on. Lay off points on H 6 corre- 
sponding to those on F 6. 

155. Inasmuch as the point H does not move in accordance with 
the law of any of the base curves no precise computation can be 
made for the size of the pitch circle for any given pressure angle 
and it may be omitted. Instead, a minimum radius M H of the 
pitch surface may be assumed. If it is desired to control the pres- 
sure angles it may be done by first constructing the pitch surface, 
H V W, and then measuring the angles at the construction points. 
Some of these are shown in Fig. 66, at H, 3, 6, and 8, and are 20°, 
— 12°, 48°, and 57°, respectively. If these angles should prove un- 
satisfactory a larger pitch circle, or a differently proportioned rocker, 
may be used. Or, an approximate computation for radius of pitch 
circle by the method which is explained to advantage in connection 
with the next problem, paragraphs 164 and 165, may be used. 

156. To construct the second cam, take the distance A E as a 
radius and A, 1, 2 . . . as centers and mark the points E, 1,2. . . . 
Again, with the latter points as centers and E J as a radius, mark 
the points J, 1, 2 . . . and transfer these to L, 1, 2. . . . With 
the latter points marked, the pitch surface of the second cam, P Q R, 
is constructed in the same way as was the first cam. 

The angle between the keyways, marked 393/2° in Fig. 66, must 
be carefully measured and shown on the drawing. 

157. Exercise problem 18a. Required a cam mechanism that 
will draw the numeral 8, the marking point moving with uniform 
velocity. 

158. Problem 19. Cams for reproduction of handwriting. 
Required a cam mechanism to reproduce the script letters S t e. 

159. The first step in the solution of this problem is to write the 
letters carefully, for if the machine is properly designed it will re- 
produce the copy exactly as written. The copy is written at A in 
Fig. 67. 

160. The next step is to decide on the kind of mechanism and 
the type of cams to be used, for the problem may be solved by a 
number of different combinations. The mechanism for this problem 



-H^rL^ 




Fig. GG. — Problem 18, Cam for Drawing an Ellipse 




Fig. 67. — Problem 19, Cams for Reproducing Script Letters, etc. 



CAMS FOR REPRODUCING GIVEN CURVES OR FIGURES 83 



:t> 



r 



will consist of two radial single-acting cams 
mounted on one shaft, and a swinging rocker 
arm mounted on a pivot which is moved forth 
and back on a radial line as shown in Fig. 67. 
This mechanical combination is selected for this 
problem because it involves methods of construc- 
tion not used in any of the preceding problems. 

161. The actual work of construction is started 
by marking off a series of dots along the lines of 

S the entire copy, as shown at A, and marked from 
a zero to 64. Inasmuch as there is some latitude 
fi in the spacing, and consequently in the number 
s of these dots, as will be explained presently, it is 
o advisable to use a total number of dots whose 
o least factors are 2 and 2, 2 and 3, or '2 and 5. 
| This is not essential but it will facilitate the work 
> later on. 

162. The matter of placing the dots is per- 
il haps the most important item of the entire prob- 
2 lem, for on this depends the size of the roller and 

smooth action. In fact, with some methods of spac- 
ing, no roller can be used at all and a sharp V- 
edge sliding follower will have to be used if true 
reproduction is desired. 

The basic considerations in selecting the 
points are: 

First, that a point should be located at the 
extreme right and extreme left of each right and 
left throw, as at - 7, 7 - 16, 16 -20 . . . in Fig. 
67, A, and at the top and bottom of each swing, 
as at - 8, 8 - 13, 18 - 18 . . .; and, 

Secondly, that the marking point should start 
slowly and come to rest gradually on each stroke, 
considering both of the component directions of 
its motion at the same time. On account of this 
it is impossible to secure ideal conditions at all times and com- 
promises must frequently be made. For example, the component 
motions of the marking point D are: First, a horizontal one due 
to Cam No. 1 ; and secondly, a vertical curvilinear one due to Cam 
No. 2 and the rocker arm H G D. The intermediate points 0-7 on the 
upper swing of the letter S are so selected as to give increasing and 



84 ELEMENTARY CAMS 

decreasing spaces in the horizontal projections on D E, and the 
same points, together with point 8, are selected at the same time so 
as to give increasing and decreasing spaces when projected onto the 
arc D F. Each space between a pair of adjacent numbers represents 
the same time unit. On this basis the entire spacing of the copy 
is done. 

163. With each of the points in the group at A, Fig. 67, as centers, 
and with a radius, D G, mark very carefully the corresponding points 
on G L in group B. To avoid confusion it is essential here to adopt 
some method of identifying points so marked for later reference. 
A satisfactory method is shown at B, all the motions to the right 
being indicated below, and the motions to the left, above G L. 

164. The sizes of the cams are to be next computed. To do 
this select the largest horizontal space in section A. This is found 
between 56 and 57 and is equal to .46 of the unit of length that 
happened to be selected in this problem. Assuming that the marking 
point moves with uniform velocity over this distance, and that a 
pressure angle of 40° is suitable in this instance where no heavy 
work is done, the factor of 1.19 is taken from the table in paragraph 
30. Since there are 64 time units the length of circumference of 
pitch circle for Cam No. 1 will be 

.46 X 1.19 X 64 = 35.03, and the radius 5.58. 

165. Before calculating the size of Cam No. 2 the length of the 
rocker arm G H must be decided upon and this will be taken in this 
problem at 5 units, the same as the arm G D. Then the total swing 
of the follower point H will be H K, equal to D F, and the greatest 
swing in any one direction in any one time unit will be during the 
periods 10-11 and 61-62, shown at A, Fig. 67, both equal to .48 units. 
Making the same computation as for Cam No. 1, 

.48 X 1.19 X 64 _ 

3.14 X 2 - 5 '* Z 

equals the pitch radius of Cam No. 2. 

166. The position of the cam shaft relatively to the pivot 
arm G depends on what is desired for the position of the arc H K 
with reference to the cam center. If it is desired that the points 
H and K shall be on a radial line from the center of the cam, which 
gives best practical average results for both in and out strokes, 
proceed as follows : Draw chord D F at A in Fig. 67 ; bisect it at 
J and measure distance G J which is 4.93 units. Then the distance 



CAMS FOR REPRODUCING GIVEN CURVES OR FIGURES 85 

GO will be the hypothenuse of a right angle triangle of which one 
side is 4.93 and the other 5.82. This may be separately drawn and 
the length of the hypothenuse found graphically or it may be 
figured as follows: 



GO = V 5.82 2 + 4.93 2 = 7.03. 

167. The pitch circles for both cams may be taken in problems 
of this kind to pass through the midpoint of the total travel. Then 
M is the radius of the pitch circle of Cam No. 1 and N P the 
total range of travel of the roller center; and Ji is the radius of 
the pitch circle of Cam No. 2 and H K the total range of travel of 
the roller center relatively to G. 

168. To find points on the pitch surface of the cams proceed in 
the usual way for Cam No. 1, by dividing the circle whose radius is 
N into as many equal parts as there were dots on construction 
points at A. Draw radial lines, and on these lay off the distances 
secured from B in Fig. 67 ; for example, the distance 3 N\ is laid 
out equal to G 3. The point Ni, and other points secured in similar 
manner, will lie on the pitch surface of Cam No. 1. 

169. The construction of the pitch surface for Cam No. 2 is 
different from that of Cam No. 1, and is different also from anything 
done in the preceding problems. In this case the resultant motion 
of the arm G H is made up of rectilinear translation and rotation and 
both components must be considered in laying out the pitch surface, 
for example, as follows : With G H as a radius and point 4 of B as 
a center draw an arc intersecting the horizontal line through H at 4- 
Then when G is moved to 4 by Cam No. 1, H would be at 4 if the 
rectilinear component motion due to cam No. 1 were the only one 
acting. During the period represented by G 4, however, Cam No. 2 
must move the rocker arm through an arc Q 4, shown at A, and this 
arc must now be laid off at 4 R> The point R is then revolved to 
its proper position at T as follows: Divide the circle OG into 
sixty-four equal parts. This is readily done in this problem because 
G is taken on the same radial line with N and the radial divisions 
already made on the circle having N f or a radius need only be 
extended. Lay off the distance G 4 at 4 S. With S as a center and 
G H as a radius draw the arc 4 T. Then T will be a point on the 
pitch surface of Cam No. 2. 

170. Having determined the pitch surfaces of the two cams the 
largest possible roller for each is found by searching for the shortest 
radius of curvature on the working side of each pitch surface. For 



86 ELEMENTARY CAMS 

Cam No. 1 the size of the largest roller that can be used is that of 
the circle whose center is at U; and for Cam No. 2 it is that of the 
circle whose center is at V. In order to avoid sharp edges on the 
cams, rollers slightly smaller than these circles will be used. 

171. For assembling the cams the angles between them and the 
angles for the keyways should be carefully measured and placed on 
the drawing as shown in Fig. 67. 

A front view showing the elevations of the cams, lever arm, slide, 
and plate is given in Fig. 68. 

172. Method of subdividing circles into any desired num- 
ber of equal parts. The matter of subdividing the circle having 




Fig. 69. — Method of Subdividing Circles into Any Desired Number of Equal Arcs 



radius N, Fig. 67, into sixty-four equal parts was a simple matter 
of subdivisions. If it is required to divide the circle into eighty-seven 
equal parts the work is just as simple if a proper start is made as 
follows : Let it be required that the circle B D, Fig. 69, be divided 
into eighty-seven equal parts. Find the number next lower than 
eighty-seven whose least factors are 2X2, 2X3, or 2X5. Such a 
number is 80. Assume that the circle is 6 inches in diameter; then 
the circumference is 18.84 inches and 7 / 8 7 of this is 1.516 inches, 
which is laid off to scale on the tangent at B F. With a pair of small 
dividers, set to any convenient small measuring unit, step off divisions 



CAMS FOR REPRODUCING GIVEN CURVES OR FIGURES 87 

from F to the next step beyond B. Assume that there are 11 steps 
from F to G, then go forward 11 steps on the arc to K. Divide the 
large part of the circle K D B into eighty parts by the process of sub- 
division with the dividers as indicated by the divided angles 80, 40, 
20, 4, 2, and 1, in Fig. 09. Then B II is % of K D B, or % 7 of 
the entire circle, and the length B II will go exactly seven times into 
the arc B K. In this work nothing is said of the use of a protractor 
for laying off a large number of small subdivisions on a circle, al- 
though it may be used. The process of subdivision, however, always 
using the small dividers, gives automatically remarkably accurate 
results. 



SECTION VI.— ADVANCED GROUP OF BASE CURVES 

FOR CAMS 

173. The previous sections of the book have dealt with the 
simpler base curves which are in common use, and with their ele- 
mentary application to various types of cams. In the present sec- 
tion the simpler forms of base curves are further considered, other 
forms are treated, and new ones are proposed; all are brought 
together for comparisons. 

174. Complete list of base curves. The base curves which 
have been used in the previous sections are: 

Straight Line, Figs. 22 and 78. 
Straight-Line Combination, Figs. 23 and 82. 
Crank Curve, Figs. 24 and 86. 
Parabola, Figs. 25 and 90. 
Elliptical Curve, Figs. 26 and 102. 

Other base curves which will be considered in following para- 
graphs are: 

All-Logarithmic Curve, Fig. 70. 
Logarithmic-Combination Curve, Fig. 74. 
Tangential Curve, Case 1, Fig. 94. 
Circular Curve, Case 1, Fig. 98. 
Cube Curve, Case 1, Fig. 106. 
Circular Curve, Case 2, Fig. 110. 
Cube Curve, Case 2, Fig. 114. 
Tangential Curve, Case 2, Fig. 118. 

175. Comparison of base curves, their applications, and 
their characteristic motions. Figs. 70 to 121 illustrate: 

(1) The forms of each of the base curves, Column 1. 

(2) The form and true relative size of cam, all having the same 
data, Column 2. 

(3) The velocity diagram for each cam, Column 3. 

(4) The acceleration diagram for each cam, Column, 4. 

176. The data for all of the cams and diagrams illustrated in 
Figs. 70 to 121 are as follows: 

(a) The follower to rise 1 unit in 60° turn of the cam, 

(b) " " " fall 1 unit in 60° " " " " 

(c) The follower to remain at rest for 240° turn of the cam, 

(d) " maximum pressure angle to be 30°. 






ADVANCED GROUP OF BASE CURVES FOR CAMS 89 

177. All of the cam charts illustrated in Column 1, Figs. 70 
to 121, include only the first item in the above data and they show, 
therefore, only one-sixth of their full length. In Column 2 the entire 
cam is shown in each case, and it is drawn to one-third of the scale 
used for "the chart in Column 1. 

178. Velocity and acceleration diagrams showing char- 
acteristic ACTION OF CAMS HAVING DIFFERENT FORMS OF BASE 

curves. All of the diagrams in Column 3, Figs. 70 to 121, show the 
velocity given to the follower by the cam at every instant during 
the follower stroke. In each case the length of the diagram A C 
represents the time required by the cam to turn through 60°, the 
cam shaft being assumed to be turning with uniform angular velocity. 
The numbered scale on each diagram shows the relative velocity 
given by each cam at any phase of the stroke. 

179. All of the diagrams given in Column 4 show the acceler- 
ation given to the follower by the different cams. These diagrams 
have a special interest when it is remembered that force = mass 
X acceleration, and if the mass is the same in all cases the ordinates 
of the diagrams represent the forces necessary to move the follower 
at any instant. A diagram with a distinctively long ordinate indi- 
cates that the cam will "run hard " at the phase where the long 
ordinate occurs. The scale numbers shown on the diagrams are 
based on the uniform acceleration given by the parabola cam as 
shown in Fig. 93. 

180. The characteristic actions of different cams built 
from the various base curves will be considered, in order, in the fol- 
lowing paragraphs. 

181. The all-logarithmic curve, Fig. 70, gives the smallest 
possible cam for a given pressure angle. It differs from all other cam 
curves in that it gives the maximum pressure angle all the time that 
the follower is moving, whereas the others give a maximum pressure 
angle for an instant only. One of the disadvantages of the all- 
logarithmic cam is that it causes the follower to attain nearly its 
full velocity instantaneously, and causes it to come to rest in a 
similar manner, thus giving a shock at the beginning and end of the 
stroke. This gives excessively large acceleration and retardation 
at the ends of the stroke and causes the cam to " pound " or " run 
hard " at these phases of its action. Another disadvantage is that a 
roller cannot be used with it because the pitch surface has a sharp 
edge, or angle, on the working side as shown at C, Fig. 71. The rea- 



90 



CAMS 



COMPARISONS OF CAMS FOR DIFFERENT 



< Column 1 

Cam. Charts and Base Curves 
tor One-sixth of Cam 



Column 2 
Relative Sizes of Cam/ 




B 

V 
A 








C 


30 X 

Pitch Line 'X 


^rifT 




H 


I\ J\^ 


E 
X 


r] 








* / 


— >- 





/Eig. 9i. Tangential Base Curve, Case 1 ~FlQ. Qnfit^W^ 



ADVANCED GROUP OF BASE CURVES FOR CAMS 



91 



BASE CURVES, ALL HAVING SAME DATA 



Column 3 
Velocity Diagrams 



Column 4 
Acceleration 

Diagrams 




L 



Fig- 73? 





Y eM 2 

-1 
D 








■ — 

Fta.Tf. 

ll 


r-l 


U 
Z rr 


E"-2~ ■ 


^ 


P 

c 




Fig 
F 


81. 




8 
J 



92 



CAMS 



COMPARISONS OF CAMS FOR DIFFERENT 



Column 1 

Cam Charts and Base curves 

for One-sixth of Cam 



r~ r x 



Column 2 

Relative Sizes 

of Cams 



B 




V 




^uvn 


B 


c 




Pitch Line 


«--&»*£- 


r 




Fy 






H \I K\Js* 


A\ 






R, 




A 




_J*^Tk 




-^/"jv 


- 3 V^H 











Fig. 98. — Circular Curve, Case 1 



Fig. 99, 








FG:FC::7:i 
Fig. 102. — Elliptical Curve Fig. 103. -^j- 

















D Pitch Line 






s 




H. 


27V 

84; 


— Base Curve^fe^ 


E 




r 




4 


1 A 


,. w „ co I ,v .r v 








« — 4.20- 




> 



Fig. 106. — Cube Curve, Case 1 




Fig. 114.- Cube Curve, Case 2 



VLH I. 



p 






Jf 


- IT 


c 






Pitch Line M^'fT 




Ft 






// 1/ <Q \J^ 


E 
X 


a 

Y 




/? 






, 
















~ o 07 


> 










0.0/ 




""" 
















> 





Fig* 118.— Tangential Base Curve, 

Case 2 




ADVANCED GROUP OF BASE CURVES FOR CAMS 



93 



BASE CURVES, ALL HAVING SAME DATA— Continued 

Column 3 Column 4 

Velocity Diagrams Acceleration Diagrams 




Fig. 108 




94 



CAMS 



son why a roller cannot be used under these conditions is explained 
in paragraph 59, page 37. The construction of the all-logarithmic 
cam is explained in the following paragraphs. 

182. Problem 20. Required an all-logarithmic cam causing: 

(a) The follower to rise 1 unit in 60° turn of the cam, 

(b) " " " fall 1 " " 60° " " " " 

(c) ' ' " " remain stationary for 240° turn of the cam, 

(d) A uniform pressure angle of 30°. 

183. A brief general analysis for the method of procedure 
in solving an all-logarithmic cam problem is : 

(1) To construct a logarithmic spiral having a constant normal 
angle of 30°. The spiral is shown at B H, Fig. 122, and the constant 



B ] 






dr 






i\ 


Y&fr~ 




y^ 










\ 




Pitch 


Line 




B 


v 



Fig. 70. — (Enlarged) All-Logarithmic 
Base Curve 




Fig. 71— (Enlarged) All- Logarith- 
mic Cam 



angle is noted at JDK, where D K is a radial line and D J a line 
normal to the curve. 

(2) To lay out the assigned working angle during which the 
follower motion takes place, on a piece of tracing cloth or tracing 
paper, as at b in Fig. 123. 

(3) To mark on each leg of the angle a scale to measure the fol- 
lower's motion, as at 0' M and 0' N in Fig. 123. 

(4) To lay the tracing cloth represented by Fig. 123 over the loga- 
rithmic spiral with the apex O r of the angle always at the pole O of 
the spiral, and to rotate the tracing cloth until the two legs of the 
angle cut the spiral at such points that the difference in length of 
the two legs is equal to the assigned follower motion. This is illus- 
trated in Fig. 122 where the shaded area represents the tracing-cloth 
with the assigned angle of 60° shown at b, while C minus O A equals 
the assigned follower motion of 1 unit. 



ADVANCED GROUP OF BASE CURVES FOR CAMS 



95 



(5) To mark the included part of the logarithmic spiral A C and 
use it as the surface of the cam as shown at A C in Fig. 124. 

184. The detail construction necessary to lay out the all- 
logarithmic cam for Problem 20 is as follows : Construct a logarithmic 
spiral with a constant normal angle of 30°. This may be done 
mathematically by laying off computed values which method will be 
taken up first, or it may be done graphically as will be explained later. 




A 1 

Fig. 122. — Logarithmic Curve 
Giving Constant Pressure 
Angle of 30 Degrees 



Fig. 123. — Assigned Working 
Angle, to be Drawn on 
Tracing Cloth 



In the mathematical method the first step is to solve the following 
equation : 

r 



10 °- 4343 rio 6tan ( 90 °- a ) 



where a is the assigned press are angle and b is a unit angle taken aD 
any value which may be conveniently used later in starting the 
drawing of the spiral. The values of r and r' are shown at D and 
H respectively in Fig. 122. The angles a and b are also shown. 
A convenient angle to assume for b, in general, is G0° and it is so 



taken in this problem. 



Then — equals the number whose logarithm is 



96 CAMS 

0.4343 X ^~ X 60° X tan (90° - 30°). Solving, the value of the 

logarithm is 0.2623 and the number corresponding to this is 1.83. 
Therefore, 

/ 

-=1.83 

r 

and O H , Fig. 122, is made 1.83 times D, the included angle being 
60° in accordance with the above assumption for b. The length O D 
may be taken any length in starting the construction of the spiral. 
The two points of the spiral may now be laid down as at D and H 
with O as the pole. 

185. Intermediate points on the logarithmic spiral as at G 
may be found by bisecting the angle D O H and making G a mean 
proportional between O D and O H. Then 

OD : OG : : OG : OH 

If O D is taken as 3 units, then G = V 3 X (1.83 X 3) = 4.06. 
To find points on the spiral at closer intervals bisect angle DOG 
and find the mean proportion 7 which is equal to V 3 X 4.06 = 3.52. 
To find other points outside of a given angle, such as at 5, lay off 
the angle D 5 equal to angle D 7 and make 5a fourth propor- 
tional to 7 and D as follows : 

05 : OD : : OD : 07 

Then 05 = ~ = 2.58. 

o.oZ 

If points are desired still closer together, or if it is desired to 
extend the spiral in either direction, it may be done by the above- 
described processes, or, it may be done graphically as described in 
paragraph 187. 

186. The next detail step in the solution of Problem 20 is to draw 
an angle M O ' N, Fig. 123, on tracing cloth, equal to the assigned 
angle of 60° as given at (a) in the data, and lay off a scale on each 
leg of the angle as shown. Then lay Fig. 123 over Fig. 122, ' 
always at 0, and rotate the tracing cloth until the spiral B H inter- 
cepts the lines ' M and ' N at such points that ' C ' minus 
O' A' equals the assigned follower motion which is 1 unit as stated 
at (a) in Problem 20. This occurs when Fig. 123 is at the position 



ADVANCED GROUP OF BASE CURVES FOR CAMS 



97 



shown by the section lines in Fig. 122 where A equals 1.18 and C 
equals 2.18. The intercepted part A C of the logarithmic spiral 

// 




Fig. 122. — (Duplicate) Logarithmic 
Curve Giving Constant Pressure 
Angle of 30 Degrees 



Fig. 123. — (Duplicate) Assigned 
Working Angle, to be Drawn 
on Tracing Cloth 



becomes a portion of the cam pitch surface as shown at A C in 
Fig. 124 and its distance from the center of rotation of the cam is 
the same as the distance from the spiral arc to the pole of the spiral. 




Fig. 124. — Problem 20. All-Logarithmic Cam for Assigned Data 

Other portions of the cam surface are found in a similar manner. 
As shown at A, E, and C, in Fig. 124, the pressure angle is 30° at all 
points. 



98 



CAMS 



187. Intermediate points on the logarithmic spiral may be found 
graphically, instead of by computation as given in paragraph 185, 
as follows: From any point of a straight line, Fig. 125, lay off D 
and H in opposite directions, D and H being the values ob- 
tained by computation in paragraph 184 and shown in Fig. 122. 
At 0, Fig. 125, erect a perpendicular line. Find the midpoint 0\ 
on the line D H, and with this as a center for the compass draw the 
semicircle D G H. Then G will be a mean proportional between 
D and H and may be laid out as the ordinate of the loga- 
rithmic spiral, as at G, Fig. 122, where G bisects the angle DO H. 




D s o o, o, H 

Fig. 125. — Graphical Method for Finding Intermediate Points ox Logarithmic 

Curve 



To find a fourth proportional graphically proceed as follows: 
Lay off the two known values, D and H, which are shown in 
Fig. 122, at right angles to each other as shown at D' and H in 
Fig. 125. Find the point O2 on OH that is equidistant from D f 
and H, and with this as a center draw the semicircle H D' 3, giving 
the length 3 as the fourth proportional. This latter distance is 
laid off at 3 in Fig. 122 where the angle D 3 is equal to angle 
DOH. 

188. A GRAPHICAL METHOD FOR CONSTRUCTING A LOGARITHMIC 
SPIRAL WHICH HAS A GIVEN CONSTANT NORMAL ANCLE is illustrated in 

Fig. 126. This method, referred to in paragraph 184, is based on 
the following theoretical property of the logarithmic spiral, namely, 
that all pairs of radiants having a common difference embrace equal 
lengths of arcs en the spiral. 



ADVANCED GROUP OF BASE CURVES FOR CAMS 



99 



189. The principle stated in the previous paragraph may be 
graphically applied only approximately, bul with all accessary pre- 
cision, by first drawing the lines .1/ /' and P N, Fig. L26, making the 
desired angle with each other. This angle will be 30° if a spiral having 
a constant normal angle of 30° is required, 40° if a constant pressure 
angle of 40° is required, etc. From a point 0, where the vertical 
intercept D is equal to about the estimated short radius of the cam, 
draw a series of equidistant vertical lines as at B, C, E, etc. With 
B F as a radius and as a center draw the short arc 1 ; with D F as a 
radius and D as a center draw arc 2. The intersection of arcs 1 and 2 
will give the point H on the spiral. Again, with C G as a radius and 




Fig. 126. — Graphical Method ion Constructing a Logarithmic Curve Having a 
Given Constant Normal Angle 

as a center draw arc 3; and with F G (equal D F) as a radius and H 
as a center draw arc ■ 4. The intersection of arcs 3 and 4 will give a 
second point L on the logarithmic spiral. It will now be noted that 
the two pairs of radiants H 0- DO and LO-H have a common 
difference, and that the logarithmic arcs D H and H L are equal 
(approximately), which accords with the general principle laid down 
in the preceding paragraph. 

190. To be exact, in the matter of the graphical construction of 
the logarithmic spiral, it must be noted that it is the chords from D to 
H and from H to L that are equal according to this method of con- 
struction and not the arcs as they should be theoretically; but where 
the vertical construction lines are taken close together and where the 
distance D F is, therefore, small, the error in the curve is negligible. 



100 



CAMS 



In the present case the ultimate distance R when drawn with aver- 
age care to a scale several times that shown in Fig. 126, varied from 
the computed value by less than .01 inch. The part of the curve 
from D to Q will depart from theoretical values faster than the part 
from D to R, due to the sharper curvature of D Q, but the effect 
of this may be overcome, if desired, by making the vertical con- 
struction lines to the right of D closer than those to the left oiOD. 

191. The all-logarithmic cam may be constructed by a 
purely graphical method, and without any mathematical com- 
putation whatever. In Problem 20, for example, it would only be 
necessary to follow the directions in paragraphs 188 and 189, making 
the angle a of Fig. 126 equal to 30° which is the assigned pressure 
angle in the problem. This would give the proper logarithmic curve 
identical with the one in Fig. 122. From this point on, the direc- 
tions given in paragraph 186 apply. If a pressure angle of any other 
size were desired, say 45°, the angle M P N Fig. 126, would be 
made 45°. 

192. Exercise problem 20a. Required an all-logarithmic 
cam which will cause a follower to : 

(a) Rise two units in 45° turn of the cam. 

(b) Remain stationary for 135° " " " " 

(c) Fall two units in 45° « << << « 

(d) Remain stationary for 135° " " " " 

(e) The constant pressure angle to be 30°. 

193. A logarithmic-combination cam may be used to overcome 

the disadvantages (paragraph 181) 
of the all-logarithmic cam and at 
the same time to sacrifice very little 
in the matter of increased size. This 
is accomplished by substituting 
rounded surfaces for the angular 
surfaces formed by the all-logarith- 
mic curve. When the rounded sur- 
face thus substituted is derived 
from parabolic base arcs the best 
results are obtained. A cam in 
which this has been done is shown 

in Fig. 75, where the curves A Y and 
Z C are arcs of a parabola base and the center portion Y Z is an 




Fig 



75. — (Enlarged) Logarithmic- 
Combination Cam 



ADVANCED GROUP OF BASE CURVES FOR CAMS 101 

arc of a logarithmic curve. To illustrate an actual case, a prob- 
lem having the same general data as Problem 20 will be discussed 

in the following paragraphs. 

194. Problem 21. Required a logarithmic-combination cam 
causing the follower to : 

(a) Rise 1 unit in 60° turn of the cam. 

(b) Fall 1 " " " " " " " 

(c) Remain stationary for 240° turn of the cam. 

(d) The maximum pressure angle to be 30°, and the easing-off 
base curves to be parabolic arcs. 

195. A brief general analysis of the method of procedure in 
solving problems of this kind is: 

(1) To draw a general logarithmic curve on rectangular coordi- 
nates, the longest and shortest ordinates of which will correspond to 
the estimated longest and shortest radii of the cam, or the longest 
and shortest radii of a series of cams if a series should happen to be 
under design. 

(2) To compute the length of rectangular cam chart, as directed 
in paragraph 198, and to draw the rectangle on tracing cloth or 
tracing paper. 

(3) To construct parabolic arcs within the rectangular cam 
chart as directed in paragraph 199. 

(4) To place the cam chart as now drawn on tne tracing cloth, 
over the logarithmic curve, so that the logarithmic curve will be 
tangent to the two parabolic arcs while the bottom line of the chart is 
parallel to the abscissa of the logarithmic curve. The distance 
between the bottom of the chart and the abscissa will be the shortest 
radius of the cam. 

196. The first step in the detail of the solution of Problem 21 is 
to construct a logarithmic curve on rectangular coordinates as in 
Fig. 127. This curve is a perfectly general one and if it is drawn 
with a wide enough range of ordinates will do for all possible log- 
arithmic-combination cams, independently of all specific data. To 
construct the logarithmic curve draw a horizontal abscissa line 0', 
Fig. 127, and erect a series of ordinates one unit apart as on both 
sides of r, making their length a geometrical progression. To do 
this, make the first ordinate drawn, say L, equal to 1 unit and all 
succeeding ordinates such as ri, r-2 longer than the preceding ordinate 
by using any common multiplier throughout; also, all preceding 



102 



CAMS 



ordinates such as r', r", if tney are necessary, shorter by the inverse 
of the same ratio. For example, if L equals 1 and if the common 
multiplier is taken as 1.25 (it may be any convenient number), then 
n = 1X1.25 = 1.25, r 2 = 1.25X1.25 = 1.5625. r 3 = 1.5C25 X 1.25 = 



1.953, etc.; also r' = 1 X 



_L = .8, r" = .8 X r ~ = .04 etc. 
1.2o 1.2o 



The 



lengths should be accurately computed up to the length of the 
maximum radius of the largest cam that is likely to be used and the 
curve L G carefully drawn. 





















Gy 






1 


\^ 








L^ " 
















r" 


r' 


r 


] y^ 


r 2 


r 3 






r a 


•r 7 


0" 


C 


) C\ 


B 






















Fig. 127 



-General Logakithmic Curve Showing Subtance:;t, V— ?ul in Solving 
a Wide Range of Logarithmic Cam PROBi.r.:.:^ 



197. The length of the sub-tangent, s, in Fig. 127, is next found by 



the formula, s = 



.434 

log. m 



, where m is the common multiplier used in 



laying out the logarithmic curve L G. Since the value of m is 1.25 

434 
in this problem, s = ^— = 4 .48. This value of the sub-tangent may 

also be found graphically by drawing tangents to the logarithmic 
curve, by eye, at several points and taking an average of the sub- 
tangents thus found. This average value will probably be close 
enough for most practical work. The tangent line at A is shown at 
A C, Fig. 127. The length of the sub-tangent, B C, will be the same 
for each tangent line if it is accurately drawn. 



ADVANCED GROUP OF BASE CURVES FOR CAMS 



103 



198. A special form of rectangular diagram, Fig. 128, depending 
on the data is now constructed, its length being: 



b ir s tan a 
~180 ' 



where I = length of diagram, 

b = assigned angle of action, 

s = length of sub-tangent of the logarithmic curve as found in 

the preceding paragraph, 
a = assigned pressure angle. 




Fig. 128. — Rectangular Chart Used in Design of Logarithmic-Combination Cam 



Taking the figures from the data for this problem, and the 
value of s as found and substituting in the above formula, 



1 = 



60 X 3.14 X 4.48 X .577 
180 



2.71. 



The height of the diagram is the continuous motion of the fol- 
lower in one direction and is 1 unit in this problem as indicated at 
R C, Fig. 128. Draw the rectangle, as shown at A R C at or near the 
top of a piece of tracing cloth or tracing paper, leaving a length 
under it equal at least to what the short radius of the cam is estimated 
to be. 

199. Parabolic easing-off arcs for logarithmic-combination 
cam. The length of the rectangular diagram is now divided into at 
least 8 equal parts which are sufficient for practice problems, but in 
practical applications at least 16 divisions should be taken. A 
diagram divided into 8 parts is shown in Fig. 128. Construct a para- 
bola with vertex at A and passing through the midpoint of the dia- 
gram as at P. This is done as explained in detail in paragraph 35 



104 



CAMS 



and, briefly as follows: Divide A B into a series of equal parts, the 
total number of parts being equal to the square of the number of 
construction spaces between A and J. In this problem there are 
four construction spaces and so A B is divided into 16 equal parts 
and the 1st, 4th and 9th division points are projected horizontally 
to M, N and which are points on the parabola. Construct the 
similar parabolic arc C J in the same way. 

Lay the rectangular diagram constructed as above on tracing 
cloth over Fig. 127 and manipulate it, always with the line A R, 
Fig. 128, parallel with the line 0', Fig. 127, until the logarithmic 
curve L G, showing through the tracing cloth, is tangent to the two- 
parabolic arcs. This occurs, in this problem, when A R is 1.55 units 
above 0' , and 1.55, therefore, is the shortest radius of the pitch 
surface of the cam. For precision work later on, mark the points Y 
and Z, Fig. 128, where the logarithmic arc comes tangent to the 
parabolic arcs, 



B 






P 


V S 


c 


16 






o 






a 


w 


V 


- 


a 




* 






K. 












9 II 


4 

1 


' M 


N 


i? 


" 


/ 










»1 


A 


L HXl 


J ' 


K 


R 








o 

















Fig. 128. — (Duplicate) Rectangular 
Chart Used in Design of Loga- 
rithmic-Combination Cam 




Fig. 129. — Problem 21. Logarithmic- 
Combination Cam with Parabolic 
Arcs at Ends 



200. The cam may now be constructed, drawing first the circle, 
Fig. 129, having a radius Q A of 1.55 units. Lay out the angle A Q C 
equal to the assigned 60° and divide it into equal spaces by as many 
radial lines as there are ordinates in Fig. 128. Transfer the ordinates 
L M, H F, etc., from Fig. 128 to Fig. 129 and draw the pitch surface 
of the cam through the points A, M, F, etc. The working surface 
would be a parallel curve distant from the pitch surface by the radius 
of the follower roller. With this cam there would be uniform accel- 
eration of the follower from A to Y where the pressure angle reaches 



ADVANCED GROUP OF BASE CURVES FOR CAMS 



105 



30°. This angle remains constant until Z comes into action, when 
the follower is uniformly retarded to zero at C. 

201. If it is desired to know the pitch circle of the cam it may be 
found by noting, in Fig. 128, where the logarithmic arc comes tan- 
gent to the starting parabolic arc. This is at Y and in this problem 
it is .06 unit from the bottom of the diagram. This distance is laid 
off at A S in Fig. 129 to obtain the pitch circle S T. If it is desired 
further, to obtain the cam chart which is necessary to draw the veloc- 
ity and acceleration diagrams, it may be found as represented in 
Fig. 74 where the length D F r is equal to the length of the arc S T in 
Fig. 129 when both are drawn to the same scale. D F r is the pitch 
line of the chart, and A R is .06 unit below it, this value being taken 
from Fig, 128, The length of the ordinates, LM,H F } etc., in Fig. 74 



v 





















K 


-30^ 






K 




^ 


Z 










\ 




F^ 


















A 


i/ 






Pitch 


Line 






F' 




{ 










\ 









D 

A L H I J R 

Fig. 74. — (Enlarged) Logarithmic-Combination Base Curve 



are equal to those in Fig. 128 when both figures are drawn to the same 
scale. It will be noted that no factor is given in connection with 
the cam chart for the logarithmic cam as it is for other cams. There 
is no constant factor; it varies with each problem. 

202. The rates of acceleration and retardation that will be given 
by the cam at the ends of the stroke are arbitrarily determined in 
Fig. 128 by causing the parabolic arcs to pass through P and J. 
With the parabolic arcs so taken good average results will be ob- 
tained, as compared with other small cams. If different accelera- 
tions and retardations are desired for the follower the point P may 
be located further up, or further down, and the cam will be either 
smaller or larger. 

203. Exercise problem 21a. Required a logarithmic-com- 
bination cam with parabolic easing-ofT arcs which will cause a fol- 
lower : 

(a) To rise 3 units in 90° turn of the cam. 

(b) To remain stationary for 180° turn of the cam. 



106 



CAMS 



(c) To fall 3 units in 90° turn of the cam. 

(d) The maximum pressure angle to be 35°. 

204. The characteristics of a cam having a straight base 
line have already been considered in the early part of this book, in 
paragraph 32. A sharp or V-edge sliding follower is the only kind 
that can be used with the straight base line for true results; a roller 
cannot be used for reasons explained in paragraph 59. The form of 
the pitch surface of the cam that is derived from the straight base line 
is the Archimedean spiral. The straight base line gives the smallest 
simple cam for a given maximum pressure angle. Its method 




Fig. 78. — (Enlarged) Straight Base Line Fig. 79. — (Enlarged) Problem 22. 

Straight Base Line Cam 



of construction is illustrated in Figs. 78 and 79 for a problem of the 
following data: 

205. Problem 22. Required a cam with a straight-line 
base in which the follower : 

(a) Rises 1 unit in 60° turn of the cam. 

(b) Falls 1 " " 60° " " " " 

(c) Remains stationary for 240° turn of the cam. 

(d) The maximum pressure angle to be 30°. 

206. In accordance with formula (1), paragraph 29, the radius 

1 X 1 73 
of the pitch circle will be 57.3 — — - — = 1.65 which is drawn at D 

60 

in Fig. 79. The given angle of 60° for the rise is laid off at D C 

and divided into any convenient number of construction parts, six 

being shown by the radial extension lines in the Figure. The first 

line is $ of F C, the second f of F C, etc. Inasmuch as no roller 



ADVANCED GROUP OF BASE CURVES FOR CAMS 



107 



can be used with this cam the pitch and working surfaces coincide, 
and a V-edge follower must be used for true results. The max- 
imum pressure angle occurs at the start and grows smaller towards 
the end of the stroke; in this problem it diminishes to 16° as indicated 
in the Figure. 

207. Exercise problem 22a. Required a cam with a straight- 
line base in which the follower: 

(a) Rises 3 units in 120° turn of the cam. 

(b) Falls 3 " " 120° " " " " 

(c) Remains stationary for 120° turn of the cam. 

(d) The maximum pressure angle to be 30 a . 

208. The straight-line combination base curve, Fig. 82, 
gives increasing velocity and acceleration at the beginning of the 

c 





yS\l 


.i 


Pitch Line 




F 

\ 


i 

-f 




E 




' 


^'n X 


R 















~T' 



Fig. 82 (Enlarged) Straight-Line Combination Base Curve 

stroke, uniform velocity and zero acceleration during a large middle 
portion of the stroke, and decreasing velocity and retardation at the 
end. The length of the period for uniform velocity and the amounts 
of acceleration and retardation depend entirely on the length of the 
easing-off radius. This may be taken at w 

any value. The acceleration diagram in 
Fig. 85 is based on a radius equal to the 
follower motion as shown at B A, Fig. 82. 
The shorter this radius is taken, the nearer 
the straight-line combination curve ap- 
proaches the cam having a straight base 
line, Fig. 78, and the action at the beginning 
and at the end of the stroke becomes more 
violent. The longer the easing off radius is 
taken, the nearer the combination curve approaches the circular base 



ii 



Fig. 85. — (Duplicate) Accel- 
eration Diagram for 
Straight - Line - Combina- 
tion Cam 



108 



CAMS 




Fig. 89. — (Duplicate) Ac- 
celeration Diagram for 
Crank Curve Cam 



curve of Fig. 98 and the smoother the action will be, but in this case 
the cam will be relatively large. The combination curve cannot 
be laid out directly on the cam itself; the chart must be constructed 
first and the ordinates transferred to the cam drawing. The con- 
struction of a cam from the combination curve is illustrated in 
Problem 10, page 55. 

209. The crank curve base, Fig. 86, described in paragraph 34, 
gives increasing variable velocity during the first half of the stroke 

and decreasing variable velocity during 
the last half. The acceleration and re- 
tardation are also variable, being greatest 
at the ends as may be noted by an in- 
spection of Fig. 89. The suddenness of 
the starting action compares with that of 
a body starting to fall under the action of 
gravity, approximately as 1.23 is to 1.00. 

210. The crank curve is sometimes called the harmonic curve 
due to the fact that it gives to the follower a motion similar to that 
described by the foot of a perpendicular let fall on the diameter of a 
crank circle from a crank pin moving with uniform velocity in that 
circle; or, in other words, a motion similar to that of a crosshead 
which is operated from a uniformly rotating crank with a T-headed 
or " infinite " connecting rod. It will also be observed that the 
crank curve is a projection of a helix onto a plane surface parallel to 
the axis of the helix, and is, further, a sine curve, or sinusoid, in which 
the length or pitch is not necessarily equal to the circumference of 
the construction circle. 

211. Effect of crank curve following its tangent line 
closely. The crank curve has the marked characteristic, under 
ordinary conditions, of following its tan- 
gent so closely, as, for example, on each 
side of E, Fig. 86, that when the crank 
curve chart is bent to form the cam, as 
explained in paragraphs 54 and 55, a j^g 
maximum pressure angle slightly greater 
than 30° is produced in the cam. In the 

case illustrated in Fig. 87 the pressure angle would still be 30° 
at E but it would be 30° 27' just to the left of E towards A . If it were 
desired to keep the maximum pressure angle exactly 30° instead of 
30° 27', it could be done by moving all the points from A to C, 




. — (Duplicate) Crank 
Chart Curve 



ADVANCED GROUP OF BASE CURVES FOR CAMS 



109 



Fig. 87, outward radially by the amount d given in the following 
formula : 

.5h 



d = 



V 



1 + jr, cot 2 a 




Fig. 87. — (Enlarged) Crank Curve Cam 

where d = distance the points on the pitch surface, as obtained in 
the ordinary way, would have to be moved out radially 
to obtain exact size of crank curve cam for a given max- 
imum pressure angle. 

h = total rise of follower. 

b = angle turned by cam during the follower's total rise, in 
radians. If b is taken in degrees the number 180 must 
be used in place of tt. 

a = pressure angle in degrees. 

The maximum pressure angle of 30° would then occur where 
the enlarged pitch surface crosses the pitch circle which would be 
slightly to the left of E, Fig. 87. The cam would be .09 larger in 
maximum radius, or 3.19 units from to C instead of 3.10 as shown 
and as used in practice. 

212. Another way of obtaining exact results with the crank curve 
would be to compute the length of the chart from the following 
formula : 



Z = .5 6 



>w 



7T 2 
1 + 7^ COt 2 



110 



CAMS 



For the case in hand I would equal 2.77, which it will be noted is .05 
larger than the practical value used in Fig. 86. With this length of 
chart the crank curve base line would not reach a 30° angle in the 
chart but the cam pitch surface would, at a point just inside of the 
pitch circle. 

213. Parabola. This chart curve, Fig. 90, already discussed in 
paragraphs 35 and 36, gives uniformly increasing velocity to the 



B 
G 

■4-\. 














c 






— 7 Pitch \ 


yr lime 






J 

F y< 


-jj==a. 


H 


I \.J^ 


E 
X 


R " 


■ A 




s^N 




Q 

















Fig. 90. — (Enlarged) Parabola Base Curve 

follower up to mid stroke when the velocity is twice that produced 
by the straight base line as illustrated in Figs. 92 and 80, respectively. 
The follower has uniformly decreasing velocity during the second 
half of its motion. Both the acceleration and the retardation are 
uniform throughout the entire stroke as shown by the horizontal 
lines BD and FH in Fig. 93. 



i» 




Fig. 80.' FiQ. 92. 

Fig. 80. — (Duplicate) Velocity Diagram for Straight Base Line Cam 
Fig. 92. — (Duplicate) Velocity Diagram for Parabola Cam 
Fig. 93. — (Duplicate) Acceleration Diagram for Parabola Cam 



214. Perfect cam action. The parabola is the only base curve 
that gives a theoretically perfect motion so far as inherent smooth- 
ness of action is concerned. It gives to the follower the same gentle 
motion on starting as a falling body has when starting from rest, and 
it brings the follower to rest at the end of its stroke with the same 
gentle action reversed. For this reason the curve is sometimes called 
the " Gravity Curve." The curve for the parabola cam is also 
referred to by some as the curve of squares from the fact that one set 
of ordinates of the curve vary as the square of the time, as may be 
noted from the fact that the construction numbers 1, 4, 9, and 16 in 
Fig. 90 are the squares of 1, 2, 3, and 4, respectively. In Fig. 106 



ADVANCED GROUP OF BASE CURVES FOR CAMS 111 

which will be described later, a curve is used in which the ordinates 
of the curve vary as the cube of the time. 

215. The parabola base curve will also operate a follower 
with the least amount of effort of any of the base curves, due to the 
fact that the acceleration is constant. Since the mass is also constant 
in cases under comparison, the force required to move the follower 
will be constant and may be represented by 1.0 as shown in Fig. 93 
in comparison with a maximum of 1.8 for the logarithmic-combina- 
tion cam, Fig. 77; 2.0 for the straight-line combination curve, Fig. 85; 
1.2 for the crank curve, Fig. 89; 1.6 for the tangential curve, Case I, 
Fig. 97; 1.5 for the circular curve, Case I, Fig. 101; and 1.7 for the 
elliptical curve, Fig. 105. These figures are for symmetrical chart 
curves. Among the unsymmetrical chart curves shown in Figs. 110, 
114, etc., much larger direct forces even may be required to operate 
the cam as illustrated by the relative maximum values of 2.9 for the 
circular curve, Case II, Fig. 113; 4.8 for the cube curve, Fig. 117; 
and 6.4 for the tangential cam, Case II, Fig. 121. 

216. Comparison of parabolic and crank base curves. While 
the parabola base curve combines the two highest theoretical con- 
siderations, namely smoothest possible motion and least power for 
operation, it has not become so widely used as the crank curve. 
This may be due to the experience of builders of cams who have 
found that the crank curve permits of a smaller cam for a given 
pressure angle than does the parabola; or for the same size cams 
the pressure angle is the smaller for the crank curve and, therefore, 
does not " stick " or "run hard " so much as the parabola cam of 
equal size. Figures on which the above state- 
ments are based may be seen in Fig. 87 where 
it is shown that a maximum radius of 3.1 
inches is required for a lift of 1 inch in 60° 
with a maximum pressure angle of 30° when 
the crank curve is used; while in Fig. 91 a 
parabola cam is shown to require a maxi- 
mum radius of 3.8 inches for the same data. 
The crank curve has obtained some undue 
comparative credit over the "parabola" curve FlG - 91.— (Enlarged) Parabola 
on account of the fact that the " parabola" 

was constructed with spaces in some other ratio than 1, 3, 5, etc. 
While, for example, a true parabola may be constructed with 
spaces of 1, 2, 3, instead of 1, 3, 5, as used in paragraph 35, 




112 



CAMS 



the parabolic curve of the cam surface in the former case will 
not be tangent to the circular part of the cam surface, or, in 
other words, the base curve E A in Fig. 90 will not be tangent to the 
horizontal base line of the chart at A but will intersect it at that 
point. A " parabola" cam, therefore, with ordinates that are in 



B 
G 
















( 


-* 






^— 5 

-*- 7 r;tch 


A^° 


>^ Line 






f 

F ~ 


H 


H 


I \, 


7^ 


E . 

X 


R ' 


' 


A 




/^N 






Q 



















Fig. 90. — (Duplicate) Parabola Base Curve 

any other ratio than 1, 4, 9, etc., will naturally show " bright spots " 
and rapid wear at the beginning and end of the parabolic surface, 
and this has actually been erroneously charged against the true 
practical parabola cam. 

217. A further comparison of the parabola and crank base curves 
shows that their velocity and acceleration lines, Figs. 88, 89, 92 and 
93, do not differ in their maximum values to such an extent, as to 




RP 



Fig. 88. — (Duplicate) Velocity Diagram for Crank Curve Cam 
Fig. 89. — (Duplicate) Acceleration Diagram for Crank Curve Cam 




~t-'z --^ — 



R 
Tig. 93. 

Fig. 92. — (Duplicate) Velocity Diagram for Parabola Cam 
Fig. 93. — (Duplicate) Acceleration Diagram for Parabola Cam 



make a noticeable difference in the action in many cam applications, 
particularly where the smoothest motion is not essential nor where 
there is a surplus of driving power. Furthermore, the drawing of the 
crank curve has appeared to some builders as a much easier and 
better-understood procedure and this has accounted some for the use 



ADVANCED GROUP OF BASE CURVES FOR CAMS 113 

of the crank curve. It may be observed, however, that the parabola 
is really no more difficult to draw than the crank curve, and when it 
is fully understood it is quite certain that the parabola cam will come 
into a more general use in all cases except where space is extremely 
limited, or where special considerations of the follower motion as to 
spring or gravity action or as to low striking or seating velocity, etc., 
become especially desirable. The subjects of spring action and low 
striking velocities will be treated in paragraph 273, et seq. 

218. Tangential base curve. This base curve differs from the 
others in that it cannot be readily used to construct the cam. The 
cam itself is drawn first by using straight lines as the side boundaries 
of the cam lobe, the straight lines being rounded off at the ends by 
arcs of circles or other smooth curves as shown in Fig. 95. At the 
inner ends, the straight lines are tangent to a circle which has the 
center of rotation of the cam as its center. The base curve for this 
cam is useful only where it is desired to find graphically the velocity 
and acceleration diagrams, and when it is so used, it must be derived 
from the cam drawing as explained in paragraph 225. The tangential 
cam is perhaps the easiest of all cams to draw when one is not par- 
ticular about the maximum pressure angle, but it is apt to give the 
highest velocities and the greatest accelerations of all the cams when 
it is laid out "by eye " by an inexperienced person. To keep the 
tangential cam under control when being designed, requires either a 
preliminary graphical construction, or a series of computations by 
means of formulas which will give results that may be laid out 
directly. 

219. Problem 23. Tangential cam, case I. Required a tan- 
gential cam in which the follower : 

(a) Rises 1 unit in 60° turn of the cam. 

(b) Falls 1 " " 60° "' " " " 

(c) Remains at rest for 240° turn of the cam. 

(d) The maximum pressure angle to be 30° and the end of the lobe 

to be rounded off by a circular arc. 

Find: The shortest radius of pitch surface of cam, the length of 
the straight-line portion of the cam lobe, the radius of the rounding 
off curve at the end, and the largest size roller that may be used. 

220. The graphical method of construction for the tangen- 
tial cam is as follows: In a preliminary and separate drawing, con- 
struct an angle AO E, Fig. 130, equal to the given pressure angle; 



114 



CAMS 



draw a line A E at right angles to A at any distance out, and con- 
tinue A E until it intersects E; draw an angle A C equal to the 
assigned angle of action; drop a vertical line from E to C; draw 
the arc E C with L as a center ; draw the arc C G with as a center, 
and measure the distances G A and A 0. Then G A : h : : A : s, 
where h is the assigned motion of the follower and s is the correct 
radius at which to draw the line A E in the direct drawing of the cam. 




Fig. 130. 



-Tangential Cam, Preliminary Sketch in Graphical Method of Con- 
struction for Definitely Assigned Data 



In the present illustration G A, Fig. 130, is 1.33 units and A is 4 
units. Therefore, in the direct drawing of the cam, Fig. 95, 

h X A = 1 X 4 
33 



GA 



1 



3.00, 



an.d this value is laid off at A Fig. 95. The pitch surface of the 
cam A E C is then drawn by repeating the operations in precisely the 
same order as in the preliminary drawing described above. The 
maximum pressure angle will be 30° at E where the circular easing-off 
arc is tangent to the straight line. The maximum radius of the 
roller would be E L, but as this would leave a sharp edge on the 
working surface of the cam, a value of z /± E L is taken as the radius, 
thus giving W N P as the working surface of the cam. 



ADVANCED GROUP OF BASE CURVES FOR CAMS 



115 



221. Analytical method of construction of the tangential 
cam. A direct drawing of the tangential cam may U 4 made from 
value obtained from a series of formulas having the following nota- 
tion, in which all linear dimensions are in inches and all angular 




Cir^ e 



Fig. 95. — (Enlarged) Problem 23. Tangential Base Curve Cam, Case 1 

dimensions in degrees unless otherwise specified. All symbols are 

illustrated in Fig. 131 which is for a general case: 

h = total motion of follower. 

x = fraction of follower's motion while rolling on the straight sur- 
face of the cam, or, fraction of stroke during which acceleration 
takes place. 

a = maximum pressure angle. 

b = time allotted by the data to the follower motion, measured in 
angular motion of the cam in degrees. 

s = radius of pitch surface to which the straight pitch line is drawn 
tangent. 

t = length of straight edge of cam on both pitch and working surface. 

p = radius of pitch circle. 

d = largest radius of pitch surface of cam. 

c = angle turned through by the cam when the full motion of the 
follower is reached, c will equal b when the straight part of 
the cam is not assigned in the data. 



116 



CAMS 



e = radius of circular arc for rounding-off outer corner of pitch cam. 
r = radius of roller. 

w = radius of working surface to which the straight working line of 
the cam is drawn tangent. 




Fig. 131. — Tangential Cam, Showing Terms Used in the Direct Construction bt 
the Analytical Method 



222. When the length of the straight part of the cam is not 
assigned in the data, c and b will be equal. When the length of the 
straight part is assigned c will figure out differently from b; if it 
comes less the problem is possible with the assigned data; if more, 
the length of the straight part must be reduced. 

The general formulas are : 



s = 



x h 



sec a 



. • (1) 



t = s tan a 



(2) 



sin a 



cctc = ifi~S ' 



r < e 



(3) 

(5) 
(7) 



s + h 



(4) 



e = d 



sin c 



w = s—r 



. . . (6) 

• • . (8) 



ADVANCED GROUP OF BASE CURVES FOR CAMS 



117 



223. With the data of the present problem, equation (5) must 
be solved first, for it is the only one in which all the terms but 
one are known. This formula is solved for t. With t known, formula 
(2) may be solved for s, then formula (1) for x, and so on in order 
with equations (3), (4), (6), (7), and (8). These formulas give the 
following values in the present problem: 



t =1.73 


5 =3 


x = .46 


V = 3.46 


d = 4 


e = 2 


r = 1.5 


w = 1.5 



224. With the above values, the cam in Fig. 95 is laid out in the 
following manner: Lay off given angle of 60° at DOC, draw circle 




Fig. 



-(Duplicate) Problem 23. Tangential Base Curve Cam, Case 



having radius A equal to s, draw straight part of cam A E equal to t, 
draw circular arc E C with center on C and with radius of L C 
equal to e, call r = .75e and make A W equal to it. Then W N P is 
the working surface of the cam where A W is the radius of the roller. 
The length W N of the straight part of the working surface is the 
same as the length of the straight part of the pitch surface, and the 
circular arc N P of the working surface has the same center as the 
arc E C of the pitch surface. The values d and w are automatically 
included in the process of the above described layout. 




118 CAMS 

225. If it is desired to construct the cam chart, Fig. 94, for the 
tangential cam in order to find the velocity and acceleration diagrams, 
the pitch circle of the cam, Fig. 95, should be drawn with the radius 
equal to E as computed above, and radial intercepts should be 
placed at regular distances as shown at H, I, etc., in Fig. 95. Then 

draw part of the cam chart with 
length equal to pitch arc D F, 
when both are to same scale, and 
with height equal to h. Draw 

Fig. 94.-(Duplicate) Tangential Pitch line DF Oil the chart at 

Base Curve, Case i a distance above A R equal to 

D A on the cam when both are 
to the same scale. In general the pitch line on the chart will 
not be half way up, although it appears so in this problem. Take 
the lengths of the radial lines at H, I, etc., which are shown on 
the cam in Fig. 95 and lay them off at equally spaced distances on 
the chart, Fig. 94, and draw the chart base curve A E C through the 
extremities of these lines. 

226. The tangential cam for this case has a characteristic 
retardation curve in that it is convex downward as shown from F 
to H in Fig. 97, while the retardation curves for all other cams that 
have intermediate maximum ordinates are either straight or con- 
cave. This characteristic may be an advantage in some cam appli- 
cations and will be referred to in paragraph 273 et seq. on the use of 
springs for returning the follower. The pressure angle factors for 
this curve, for the data given in this problem, are: 5.28 for 20°, 3.62 
for 30°, 2.82 for 40°, 2.36 for 50°, and 2.09 for 60°. These factors 
are used for the ordinates of curve No. 9 in Fig. 132 which shows that 
the tangential cam, for the data of Problem 23, has the advantage 
of smaller size over the parabola, circular, elliptical and cube cams 
when the lower range of pressure angles are used, but that it begins 
rapidly to lose this advantage at angles of about 36°. 

227. Further characteristics of this tangential cam that 
may be used to advantage in assigning data, are that if the angle 
turned through by the cam is twice the pressure angle, the maximum 
retardation for the circular easing-off arc of the cam will occur at 
the end of the stroke as shown at C H, Fig. 97 ; and that the retarda- 
tion at the point on the cam where the arc joins the straight line will 
be, .866 C H as shown at E F, Fig. 97. If the angle turned through 
by the cam during the motion of the follower is greater than twice 



ADVANCED GROUP OF BASE CURVES FOR CAMS 



119 







D 


r* 




77 






1 




A 


E 


2 




6' 


***-«. 


'fe 


— \ C 

r 


Fig. 97.— 


(Duplicate) Accel- 


E RATION 


Diagram for 


Tan- 


GENTIAL 


Cam 







the pressure angle the retardation value will still be a maximum at 
the end but will be less than .866 of this value at the point where 
retardation begins, that is, E F will be still shorter in comparison with 
C H than it is shown in Fig. 97. This 
condition has the practical value in that it 
allows a lighter-weight, or smaller spring 
to return the follower where a spring is 
used. If the angle turned through by the 
cam during the motion of the follower is 
less than twice the pressure angle the re- 
tardation at E F will be greater than .866 
C H, and if it is much less the retardation 

value will be a maximum at the point where the easing-ofT arc joins 
the straight line, that is, E F will be greater than C H. 

228. Exercise problem 23a. Tangential cam, case I. Re- 
quired a tangential cam in which the follower : 

(a) Rises lJ/2 units in 50° turn of the cam. 

(b) Falls 13^ units "50° " " " " 

(c) Remains at rest for 260° turn of the cam. 

(d) The maximum pressure angle to be 30°, and the end of cam 
lobe eased off by a circular arc. 

229. Circular base curve, case I. This curve, Fig. 98, is 
made up simply of two equal circular arcs as shown at A E and E C. 




Fig. 98. — (Enlarged) Circular Base Curve, Case 1 

It is the limiting case of the straight-line combination curve in 
which the two easing-ofT arcs are so large as to meet and eliminate 
the intermediate straight line entirely. The circular base curve 



120 



CAMS 



gives variable velocity and acceleration to the follower the first half 
of the follower stroke, and also variable velocity and retardation 
during the last half, as shown in Figs. 100 and 101. It will be noted 




rpD 


-1 




A— J&-K 






C 


ik ^~. 


^ 


_ 


II 


^^F 








E^^^ 


— 2— 








P 



Fig. 101 

Fig. 100. — (Duplicate) Velocity Diagram for Circular Base Curve Cam 
Fig. 101. — (Duplicate) Acceleration Diagram for Circular Base Curve Cam 

that the circular curve, and the elliptical curve shown in Fig. 102, 
give nearly the same sized cams and that the velocity and acceleration 
diagrams for each are quite similar. With the circular base curve, 
the radial distances on the cam at D, H, I, J, Fig. 99, cannot be found 




Fig. 99. — (Enlarged) Problem 24. Circular Base Curve Cam, Case 1 



directly except by means of the chart or by computation. For 
graphical construction it is necessary to draw the chart, Fig. 98, 
first and it is then a simple matter to transfer the ordinates at H, I, J, 
to Fig. 99. The length of the chart for a maximum pressure angle 
of 30° is 3.73 times the motion of the follower. 

230. The length of radius for the equal arcs in the circular base 
curve is 3.73 times the follower motion for a 30° maximum pressure 
angle. To find the length of radius for any other maximum pressure 
angle, use the formula: 

h 

r ~ 2(1 - cos a)' 



ADVANCED GROUP OF BASE CURVES FOR CAMS 

where r = the desired radius, 

a = the desired maximum pressure angle, 
and h = the given follower motion. 

Table for Circular Base Curve 



121 



For Maximum Pressure 
Alible of 


Radius of Arc is 


20° 
30° 
40° 
50° 
60° 


8.29/i 
3.73 h 
2.14 A 
1.40 h 

1.00 h 



231. Problem 24. Required a circular base curve cam 
that will cause the follower to: 



turn of cam. 

a it it 



(a) Rise 1 unit in G0° 

(b) Fall 1 " " 60° 

(c) Remain stationary for 240 

(d) With a maximum pressure angle of. 30 



O t I it It 

o 



232. The general description of the circular base curve given in 
the two preceding paragraphs will doubtless give all the necessary 
information for the solution of this problem so that only a brief 
order of procedure will be given here. The total length of chart is 

1 X 3.73 X ^ = 22.38. 

One-sixth of this length is shown in Fig. 98. The radius of the cir- 
cular arc A E, which is the same as E C, is 



1 



2Q - cos 30°) 2(1 - .866) 



= 3.73. 



Draw eight equally spaced ordinates as at //, /, J, etc., Fig. 98. The 
radius of the pitch circle of the cam is, 



22.38 



2 X 3.14 



= 3.56, 



122 



CAMS 



as drawn at D in Fig. 99. Divide the assigned arc of action D F, 
which is 60°, into eight equal parts as at H, I, J, etc. On the radial 
lines at each of these points lay off the corresponding ordinates from 
H, I, J, etc., in the chart, Fig. 98, thus obtaining the pitch surface 
A EC, Fig. 99. 

233. In some cases it may happen, when the circular base curve 
is assigned, that the length and height only of the rectangular chart 
enclosing the circular curve will be known and it may be desired to 
compute the radius and the pressure angle for the circular arc that 
must be used. For example, in Fig. 98, assume that A R and R C 
are the only known values and it is desired to find the proper radius 
of the arc EC and the pressure angle that will exist at E. The 
radius may be readily computed by simple geometry, for, the two 




s s 

Fig. 98. — (Duplicate) Circular Base Curve, Case 1 



triangles C F E and C T S will be similar in all cases and, therefore, 
SC : EC : : TC : FC. Since E F and F C are equal to one- 
half of A R and R C, respectively, their values are known and 
EC = V EF 2 + FC 2 . The length of T C is one-half of E C. 
The radius of the circular arc will be 



SC 



EC X TC 
FC ' 



234. In order to obtain the pressure angle, for the case given in 
the preceding paragraph, simple trigonometry is required, and in 
using the trigonometry, the length of the radius may also be obtained 
even more readily than by geometry. The method is as follows: 



ADVANCED GROUP OF BASE CURVES FOR CAMS 123 

In Fig. 98 the angles C S T and EST are each equal to one-half the 
angle C S E which is the pressure angle and is designated by a in 
the following formulas. The triangles C E F and C ST are similar 
in all cases. Therefore, a may be found by the following formula : 

1 CF 

i ™2 a = YF 

With a known, the radius of the arc E C may also be found as 

follows: 

E F 

ES = =-?- = C S. 

sin a 

235. Exercise problem 24a. Required a circular base curve 
cam which will cause the follower to : 

(a) Move out 3 units in 90° turn of the cam. 

(b) Remain stationary for 195° " " " " 

(c) Move in 3 units in 75° " " " " 

(d) With a maximum pressure angle of 40°. 

236. Elliptical base curve. The elliptical base curve gives 
variable velocity and variable acceleration to the follower. By using 
different ratios for the horizontal and vertical axes of the ellipse on 
which the curve is based, the velocity of the follower may be made 
to increase rapidly or slowly at the start, and the cam may be made 
small or large and still not exceed a given maximum pressure angle. 

237. Elliptical base curve, ratio 7 to 4. As stated in the pre- 
ceding paragraph the elliptical cam may be based on ellipses having 
various proportions between their major and minor axesr When the 
proportions are as 7 : 4, as in Fig. 102 where F G = 7 and F C = 4, 
the length of the chart will be 3.95 times the travel of the follower for 
a maximum pressure angle of 30°. The cam will be larger, but the 
velocity of the follower will be less at starting and stopping and 
greater at midstroke than for any of the cams described thus far. 
If a still lower starting and stopping velocity is desired with an 
elliptical cam, it may be obtained by making the ratio of horizontal 
to vertical axes on the chart as 8 : 4, 9 : 4, or greater, instead of 
7 : 4 as here used. The drawbacks to increasing the ratios above 
7 : 4 are increased size of cam and high velocity at midstroke for a 
given pressure angle. 



124 



CAMS 



238. Elliptical base curve, ratio 2 to 4. The cam produced 
from the elliptical base curve is shown, in the preceding paragraph, 
to give a certain characteristic action to the follower when the ratio 





FG:.FC:.:.1A 
Fig. 102. — (Enlarged) Elliptical Base Curve 

of the horizontal axis to the vertical axis is 7 to 4. When the ratio 
is 2 to 4, a totally different characteristic follower action is obtained 
as may be determined by a process of construction similar to that 
shown in Figs. 102 and 103. The cam itself, with a ratio of 2 to 4, 

will be much smaller for a given 
pressure angle, as may be seen by 
comparing the abscissae of curves 5 
and 11 in Fig. 132. Where it is 
desired to use a very small cam for a 
given pressure angle, the 2 : 4 ellip- 
tical curve will have an advantage 
over the ordinary straight-line com- 
bination curve above 27° as may be 
noted from an inspection of curves 
5 and 6, Fig. 132; but it is at a 
disadvantage compared with the log- 
arithmic-combination cam at all pressures angles as is shown by a 
comparison of curves 2 and 5. 

239. Elliptical base curve may be made equivalent to 
nearly all other base curves. Since the elliptical base curve 
may be constructed with any ratio of horizontal to vertical axes, it 
has a range of usefulness over the entire field covered by all the other 
base curves except the logarithmic curve. When the horizontal axis 
of the ellipse is zero, the elliptical base curve coincides exactly with 
the straight-line base. As the horizontal axis increases in length, 
the vertical axis remaining constant, the elliptical base curve crosses 
the straight-line combination curve. When the horizontal axis of 
the ellipse equals the vertical axis, the elliptical base curve is identical 
with the crank curve. As the horizontal axis continues to increase, 
the elliptical curve approximates very closely indeed to the parabola 



Fig. 



103. — (Enlarged) Elliptical 
Base Curve Cam 



ADVANCED GROUP OF BASE CURVES FOR CAMS 



125 



when the ratio of horizontal to vertical axes is as 11 to 8. A further 
general characteristic of the elliptical curve is that the starting and 
stopping velocities grow smaller, and also the accelerations or start- 
ing and stopping forces grow smaller as the horizontal axis of the 
ellipse grows larger. 

240. Cube base curve, symmetrically applied. The cube 
base curve, Fig. 106, is similar in method of construction to the 




Fig. 100. — (Enlarged) Cube Base Curve, Case 1 



parabola base curve, the only difference being that the cubes of the 
numbers 1, 2, 3, etc., instead of the squares, are used as ordinates of 
the curve. The cube curve gives extremely low and slowly increasing 
motion to the follower at the start as may be noted by an inspection 
of the velocity curve A E, Fig. 108, which shows the distinguishing 
characteristic that the velocity curve is tangent to the base line. 
The cube curve is the only one that gives uniformly increasing accel- 
eration to the follower, starting from zero, as indicated by the straight 





Eia, 108. 

Fig. 108. — (Duplicate) Velocity Diagram for Cube Cam 
Fig. 109. — (Duplicate) Acceleration Diagram for Cube Cam 



inclined line A D in Fig. 109. The disadvantage of the cube curve, 
however, is that it gives an extremely large cam for a given maximum 
pressure angle, if it is used in the same way that the preceding curves 
are used, that is, if it is made up of two similar arcs placed in reverse 
order. If the cube curve were so drawn it would be made up of 
two arcs similar to A E, Fig. 106, and the pressure angle factor would 
be 5.20 as compared, for example, with 3.46 for the parabola, and the 
maximum radius of the cam would be 5.47 against 3.80 for the para- 
bola. Because of the similarity of method of construction of the 



126 CAMS 

cube curve and the parabola, and because the large size of the sym- 
metrical cube cam renders it impractical for most cases, its drawing 
will be omitted, and instead, a modified and more practical con- 
struction of the cube cam will be illustrated and explained in the 
following paragraphs. 

241. Cube base curve unsymmetrically applied for best 
advantage. This modified cube curve will be referred to as cube 
curve, case I. Its features are that it retains the very low starting 
values of the regulation or symmetrical cube cam, and at the same 
time keeps down the size of the cam by using che regulation cube 
curve for the first half of the follower's motion and then using a 
short arc of another cube curve for the retardation in such a way that 
the maximum acceleration and retardation values shall be equal. 
In order to use this base curve several formulas are necessary and 
they, together with their notation, are given in the following par- 
agraph. 

242. Notation and formulas for cube curve cam, case I : 
h = distance moved by the follower. 

a = pressure angle. 

I = length of part of cam chart corresponding to follower's motion. 

x = length of cam chart during which acceleration takes place. 

Xi, 0:2 . . . = arbitrary lengths of cam chart taken for purposes of 

constructing chart base curve. 
2/i, 2/2 • • . — length of ordinates of cam chart corresponding to the 

values of x\, X2 . . . . 
r = radius of pitch circle of cam. 

b = angle turned through by cam in degrees during follower's mo- 
tion. 

The general formulas are: 
I = 2.427 h cot a ... (1) x = .618 I (2) 



-■(f 



y = — 7=^ — from zero to x (3) 

2V5-4 



»(i)-(i)'- (Vif - ,) 



V = h — — — 7= from x to I (4) 

180 1 ,.. 



ADVANCED GROUP OF BASE CURVES FOR CAMS 



127 



243. Problem 25. Cube curve cam, case I. Required a cube 
curve cam with unsymmetrical cube curve arcs in which the follower 
shall : 

(a) Rise 1 unit in 60° turn of the cam. 

(b) Fall 1 " " 60° " " " " 

(c) Remain stationary for 240° turn of the cam, and 

(d) The maximum pressure angle shall be 30°. 

Substituting the values given in the data in the formulas in the 
preceding paragraph, I = 4.20, x = 2.60 and r = 4.0. With these 
values, the rectangle A B C R, Fig. 106, for the cam chart may be 



2T 



D Pitch Line 




pv, 


«~^ 




S 




I 


P.nsn pil-VO 1- *" 


K^ 




E 
X 




T 




RV 


< H 7 rn l N 


J 








60 > 




^ 2.60 
























" 





Fig. 106. — (Duplicate) Cube Base Curve, Case 1 



drawn, A R being made equal tol,AX equal to x, and R C equal to h. 
The curve A E may be drawn graphically by dividing A X into four 
equal parts, A D into four unequal parts, as shown in Fig. 106, and 
projecting the division points until they meet, as at K. A D, which 
is one-half of A B, is divided into the four unequal parts as follows : 
Draw a straight line A G in any convenient direction about as shown ; 
make its length 64 units according to any convenient scale; with the 
scale still in place mark the 1st, 8th and 27th division points on A G 
and from each of these points draw lines parallel to G D until they 
intersect the side A D of the rectangle; from the latter points draw 
horizontal lines until they intersect their corresponding ordinal cs, 
as at K. Or, the values of these ordinates, as at J K, may be com- 
puted by formula (3) of the preceding paragraph by substituting the 
following values for x : x\ = }4 X , %2 = %x, x% = %x. The computed 
values of yi, ?/2, 2/3, are .008, .063, .211, respectively, and these are 
laid off at H, I, and J in Fig. 106. 

244. The portion of the cube curve from E to C, Fig. 106, is found 
by taking a series of any number of equally spaced ordinates, four 
being used in this problem and one of them marked at T S. The 
values of these ordinates are computed from formula (4) of para- 
graph 242, and arc as follows: y± = .50, y 5 = .71, y% = .87 (shown 



128 



CAMS 



at aS T), and 7/7 = .95. The corresponding values of x±, X5. . . which 
were substituted for x in equation (4) in obtaining these values were 
x± = x, X5 = x + J<£ (I — x), xq = x + Y 2 {1 — x), etc. 

245. The pitch circle of the cam is drawn with D, Fig. 107, as a 
radius and is equal to r = 4.00, obtained from equation 5. The 
values as found for the cam chart may be now transferred to the cor- 




Fig. 107. — (Enlarged) Problem 25. Cube Base Cam, Case 1 



respondingly placed radial lines from A to R, or the values as com- 
puted from formulas (3) and (4) may be laid off directly on these 
radial lines without drawing the cam chart at all. 

246. The characteristic velocities, accelerations and retardations 
produced by this case of the cube curve cam are shown in Figs. 108 
and 109, respectively. From the latter it may be seen that the 





Fig. 109. 
Fig. 108. 

Fig. 108. — (Duplicate) Velocity Diagram for Cube Cam 
Fig. 109. — (Duplicate) Acceleration Diagram for Cube Cam 



acceleration and retardation lines, A D and F H, respectively, are 
straight inclined lines, characteristic of the cube curve, as pointed 
out in paragraph 240. When the retardation line F H is extended, 
as shown by the long-dash line, Fig. 109, it passes through the zero 
point of the diagram. A cam with this characteristic may have 
particular advantages in some instances, one of which will be referred 
to later in the discussion of the relative strength of springs necessary 
\o return the follower. 



ADVANCED GROUP OF BASE CURVES FOR CAMS 



129 



247. Exercise problem, 25a. cube curve, cam, case I. Re- 
quired a cube curve cam in which the follower: 

(a) Moves up 1 unit in 50° turn of the cam. 

(b) Moves down 1 " " 50° " " " " 

(c) Remains stationary for 260° turn of the cam, and in which 

(d) The maximum pressure angle shall be 30°. 

248. Cams specially designed for low-starting velocttii;<. 
In cams where the change in velocity of the follower during the latter 
part of its travel may take place rapidly the early motion of the fol- 
lower may be made both very low and very gradual. These condi- 




Fig. 114. — (Duplicate) Cube Base Curve, Case 2 

tions as to velocity may be obtained by giving more than half the 
stroke to the acceleration of the follower, instead of one-half as has 
been the case in all preceding problems. In Figs. 110 and 114, are 
illustrated special cases of the circular and cube base curves in which 
the follower is permitted to accelerate during % of its stroke, while 
its retardation takes place in the last quarter of the stroke. In these 





Fig. 



Fig. 112. FiG. 116. 

112. — (Duplicate) Velocity Diagram for Circular Base Curve Cam, 
Fig. 11G. — (Duplicate) Velocity Diagram for Cube Cam, Case 2 



Case 2 



two cases the velocities at midstroke are approximately 1.2 and 1.0, 
respectively, as may be noted from the dash line construction in 
Figs. 112 and 116, respectively, against 2.2 and 1.7 as shown for 
similar basic curves in Figs. 100 and 108. 

249. Problem 26. Circular base curve cam, case II. Re- 
quired a cam with a circular base curve in w r hich the follower shall: 

(a) Rise 1 unit in 60° turn of the cam. 

(b) Fall 1 " " 60° " " " " 



130 



CAMS 



(c) Remain stationary for 240° turn of the cam. 

(d) Accelerate for % of its stroke, and in which 

(e) The maximum pressure angle shall be 30°. 

250. For a graphical method of construction of case II of 
the circular base curve cam, draw the cam chart as in Fig. 110 

making its length I = h X -^ X 36Q , where 



I = total length of chart. 

h = height of chart. 

/ = pressure angle factor. 

b = angle during which follower motion takes place. 



In this problem I 



1 X 3.73 X 360 
60 



22.38. 



251. One-sixth of the chart is shown in Fig. 110 at A R. Lay off 
the height A B equal to one unit and mark the point D so that 




Fig. 110. — (Enlarged) Circular Base Curve, 
Case 2 



Fig. 111. — (Enlarged) Problem 
26. Circular Base Curve 
Cam, Case 2 



AD — t, where t equals fraction of stroke assigned for acceleration. 
Draw D F. Mark the point X on A R so that A X = t X A R. 
Draw X E. Through E draw an inclined line making an angle with 
X E equal to the assigned pressure angle. Where this inclined line 
meets the lines A B and C R will be the centers for the circular arcs 
making up the base curve. These center points will be at Y and at S 
respectively. Draw the circular arcs A E and E C. Divide D E 
into a convenient number of equal parts, as at H, I . . . and draw 
ordinates to the circular arc A E. Do the same with E F. 



ADVANCED GROUP OF BASE CURVES FOR CAMS 



131 



Construct the pitch circle of the cam with a radius, 

I 22.38 



OD = 



2 X 3.14 



G.28 



3.5G, 



as shown in Fig. 111. Lay off D F equal to the assigned motion 
angle, which is 60° in this problem. The arc D F will be equal in 
length to the line D F in the chart when both are drawn to the same 
scale. Make D E on the arc equal to D E on the chart and divide 
the arc D E into the same number of equal parts as the line D E. 
Draw radial lines at the division points H, I, J, . . . and transfer 
the ordinates from the chart to these radial lines, thus obtaining the 
pitch surface of the cam from A to E. Do likewise to obtain the 
arc E C of the cam. 

252. The circular base curve, case II, gives a smaller cam 
than does case I, although both have the same pressure angle factor 
and the same chart length. The maximum radius of the cam for 




Fig. 99. — (Duplicate) Ciecular Base Curve Cam, Case 1 

case II is 3.81 against 4.06 for case I as shown in Figs. Ill and 99 
respectively. The reduction in size in case II is due to the fact that 
the pitch line D F on the cam chart is higher up in the present case, 
and, consequently, that more of the pitch surface falls inside of the 
pitch circle than in Fig. 99. The pitch circle is the same size in both 
cases. 

253. Computation for the lengths of the radii for the arcs 
A E and E C in the cam chart in Fig. 110 may be made by the fol- 
lowing formulas if desired, instead of finding them graphically as 
explained in paragraph 251. 



A Y = 



hi 



cos a 



and C S 



h(l - t) 
1 — cos a' 



132 



CAMS 



where a equals the assigned pressure angle, h equals follower motion, 
and t equals fraction of stroke assigned to acceleration. 

254. Exercise problem 26a. Circular base curve cam, 
case II. Required a cam with a circular base curve in which the 
follower shall: 

(a) Rise 2 units in 75° turn of the cam. 

(b) Fall 2 *■' il 75° " " " " 

(c) Remain stationary for 210° turn of the cam. 

(d) Accelerate for .7 of its stroke, and in which 

(e) The maximum pressure angle shall be 30°. 

255. The use of the cube curve for obtaining extremely low 
starting velocities is illustrated in Fig. 115. The cam is built up 




Fig. 114.— (Enlarged) Cube Base Curve, Case 2 
D\ H i _ 








D 


~r2 










-1 


^^-"""' 




E 




a 

Q 

u 




N 


^ M 

v. 

R 


— 1 
—2 












y^4 " 


p 






F 










G 


L-5 





Fig. 115. — (Enlarged) Problem 27. 
Cube Base Curve Cam, Case 2 



Fig. 117. — (Duplicate) Acceleration 
Diagram for Cube Cam, Case 2 



from a specially long arc of the cube base curve and it has a short 
circular base arc for easing off at the end. The chart and the base 
curve for this cam are shown in Fig. 114. The low-starting velocities 
are due to the fact that the follower has % of its stroke to reach max- 
imum velocity. This gives only \i stroke for retardation which attains 
a very high value near the end of the stroke ranging from 4.8 to 3.2, 
as shown in Fig. 117. This, of course, becomes the acceleration 



ADVANCED GROUP OF BASE CURVES FOR CAMS 133 

value at the beginning of the return stroke. Herein lies the disad- 
vantage of this cam. It is useful only where extremely slow starting 
velocity is required a1 one end of the stroke and where a rapid change 
of velocity at the other end of the stroke is immaterial. It would 
require a powerful spring to keep the follower roller in contact with 
the cam at high speeds, and if it were used on a positive drive cam 
would cause rapid wear at the beginning of the return stroke. 

256. Problem 27. Cube curve, case II. Required a cube 
curve cam with a circular arc for easing-off radius in which the fol- 
lower : 

(a) Rises 1 unit in 60° turn of the cam. 

(b) Falls 1 " " 60° " " " " 

(c) Remains stationary for 240° turn of the cam. 

(d) Accelerates during J^ of the stroke. 

(e) The maximum pressure angle to be 30°. 

257. In solving the above problem the length A X, Fig. 114, of 
that part of the chart which is given over to the cube curve is first 
found by the formula, 

3th . 

Xi = where 

tan a, 

t = the fractional part of the follower's motion devoted to accelera- 
tion. 
h = the total motion of the follower. 
a = the pressure angle. 

.ri = the length of chart under the cube curve. 
xo = the length of chart under the circular easing-off arc. 
Substituting the values given in problem 27, 

3 X .75 X 1 Q on 
xi = === = 3.90. 

.57/ 

258. The length X R of chart, Fig. 114, necessary for the easing- 
off circular arc may be computed by the formula, 

_ h(l - t) ^25 
X2 ~ tan Y 2 a " .268 ' J6 ' 

Or, the length XR may be found directly by drawing NEK 
so that it is tangent to the cube curve at E. The angle KEF will 



134 



CAMS 



then be equal to the pressure angle. Make K C equal to E K. The 
point C will then be at the end of the chart. The center for the arc 
E C will then be on the line C R extended, and at a point S which 
must also be on the perpendicular to N E K. 

259. To find points on the cube base curve A E, Fig. 114, divide 
D E into any convenient number of equal parts, six being used in the 
illustration. Draw vertical lines through each of the division points 
as at H, I, ... Draw a line A G inclining upward from A in any 
convenient direction and make the distance A G equal to the cube of 
the number of construction parts. Six parts having been chosen 
in this problem, A G will be equal to the cube of 6, or 216 units in 
length laid off to any convenient scale. At the same time lay off the 
division points 1, 8, 27, etc., which are the cubes of 1, 2, 3. etc. Draw 




Fig. 114. — (Duplicate) Cube Base Curve, Case 2 



the line G D, and then draw lines parallel to it through the points 
1, 8, 27, etc., until they intersect A D. Project these intersecting 
points horizontally until they meet the corresponding verticals from 
H,I, . . . , thus giving points on the cube base curve A E. 
260. The radius for the pitch circle of the cam will be, 



IX 360 = 4.83 X 360 
2 X 7T X b ~~~ 6.28 X 60 



4.62, 



where I = length of chart used for rise of follower and, 
b = angle during which the follower is moving. 

With the above value of r the circle through D is drawn in Fig. 
115. The arc D E F will be equal in length to the line D E F in 
Fig. 114 when drawn to the same scale, and it should be similarly 
divided and the radial lines at #, /, . . . made equal to the similarly 
lettered ordinates in the chart. The curve A E C thus obtained will 
be the pitch surface of the cam. 



ADVANCED GROUP OF BASE CURVES FOR CAMS 



135 



261. Exercise problem 27a. Cube curve, case II. Require' 
a cube curve cam with a circular easing-off arc in which the follow* 

(a) Rises 3 units in 90° turn of the cam. 

(b) Falls 3 " " 90° " " " " 

(c) Remains stationary 180° " " " " 

(d) Accelerates during .70 of its stroke. 

(e) The maximum pressure angle to be 30°. 




Fig. 115. — (Duplicate) Cube Base Curve Cam, Case 2 



262. Tangential cam, case II. The tangential cam, as stated in 
paragraph 218, is made up of straight-line sides with a circular arc 
for rounding off the end of the lobe. When the length of the straight 
surface of the cam is not specified, or when the portion of the stroke 
during which the follower accelerates is not given in the data, the 
tangential type of cam works out to good advantage. But when 
either of the above items is included in the data for the tangential 
cam it may conflict with the proper cam angle which should be 
allowed for the follower motion, as illustrated in the following prob- 
lem, which contains the same data as the two previous problems. 
The possible difficulty met with in using the tangential cam arises 
from high accelerations that may be produced. 

263. Problem 28. Tangential cam, case II. Required a tan- 
gential cam with a circular easing-off arc in w r hich the follower : 

(a) Rises 1 unit during 60° turn of the cam. 

(b) Falls 1 " " 60° " " " " 

(c) Remains stationary for 240 ° " " " " 

(d) Accelerates during J4 of its stroke. 

(e) The maximum pressure angle to be 30°. 



136 



CAMS 



264. The cam may be constructed directly by substituting values 

given in the data in the general formu- 
las given in paragraph 222, and then 
laying out the results as in Fig. 119. 
In the present problem A, Fig. 
119 equals s as found in paragraph 
a* 222, A E = t, D = p, C = d, 
angle D C = b, angle D K = c, 
and L E = e. The radius r of the 
roller and the minimum radius w of 
the working surface are not shown 
in the illustration but may be readily 
added if called for. The radius of 
the roller, however, cannot be greater 
than E L. The numerical results 
found by substituting the values given 
in the data in the series of formulas 
referred to above are as follows: 




Fig. 119.— (Enlarged) Pkoblem 28 
Tangential Base Curve Cam 
Case 2 



S = 4.84 t = 2.79 



5.58 d = 5.84 c = 39^- e = 1.47. 



265. If it is desired to construct the cam chart for the purpose of 
determining the velocity and acceleration diagrams later, it may 
readily be done: 

(1) By making the length of chart A R, Fig. 118, equal to the 
length of the arc D F on the cam drawing, 











X 


h. . 






C 




D 




Pitch Line 




X 


s^f~\ 




4 










E 
X 


G 
Y 






A 




r ' 


' 










n.-> 










2.94 




^ 








< d.Ht 

































Fig. 118. — (Enlarged) Tangential Base Curve, Case 2 



(2) by laying off the pitch line D F on the chart and subdividing 
the same as the arc D F on the cam is subdivided, 

(3) by transferring the radial lines at H, I, . . . from the cam 
to the chart and drawing them as vertical lines, thus obtaining points 
for the base curve A E K C. 



ADVANCED GKOUP OF BASE CURVES FOR CAMS 



137 



266. It will be noted that an attempt to construct a tangential 
cam in cases such as the one here represented may result in extremely 
large retardation or acceleration values, as shown in Fig. 121, the 
practical result of which will be a "hard-turning " spot at a point on 




Fig. 121. — (Duplicate) Acceleration Diagram for Tangential Cam, Case 2 



the cam corresponding to E, Fig. 119, and continuing, in lessening 
degree, to K. 

267. Exercise problem 28a. Tangential cam, case II. He- 
quired a tangential cam with a circular easing-ofT arc in which the 
follower : 

(a) Rises 2 units during 75° turn of the cam. 

(b) Falls 2 " " 75° " " " " 

(c) Remains stationary for 210° " " " " 

(d) Accelerates during .70 of its stroke. 

(e) The maximum pressure angle to be 30 °, 



SECTION VII.— CAM CHARACTERISTICS. 

268. Method of determining velocities and accelerations. 
The velocity and acceleration values in the diagrams shown in Figs. 72 
to 121 may be found by graphical methods which are simple and quite 
accurate enough for most practical purposes if precision in drawing is 
followed. The graphical method applies to all forms of cams and 
starts with the cam chart. Its application, however, is illustrated 
only in connection with the circular cam chart in Fig. 98, it being 
unnecessary to add similar lines to all the other chart drawings, as 
the constructions would be the same in every case. 

269. The use of time-distance and time-velocity diagrams. 
The chart curve A E C, Fig. 98, for our present purpose, may be 



\ 





B 
D 








\ 








C 


J 




> 

Pitch Line 


J§5 








> 


V 








H 


I K 


V 


A 








R \ 




-LI 


T 








' 


i 


^-^^N \ \ 




J- 


if 


/ 


s 




«o \ 


\ 









\ . 1 



s s 

Fig. 98. — (Duplicate) Circulak Base Curve, Case 1 

termed time-distance curve in which the abscissa A R represents 
time, and the ordinates parallel to A B represent distances traveled 
by the follower at corresponding times. If, then, the time-distance 
curve were a straight inclined line, the velocity of the follower would 
be constant. We may consider, for the instant, that the time-distance 
curve is straight at E and draw a straight line, E P, tangent at that 
point. If this were the time-distance line and if it were continued 
for a time period represented by E D, the follower would have moved 
the distance P D in the time represented by E D. If E D is consid- 

138 



CAM CHARACTERISTICS 139 

ered as a unit of time, then P D becomes a measure of velocity and its 
length is laid off in Fig. 100, at XE which is at the center of the time- 
velocity diagram. The length A C of the velocity diagram may be 
any convenient value for the purpose of comparison. The distance 
D E, Fig. 98, or one-half the length of the cam chart, was selected as 
a time unit because it is a convenient length and because the length 
of one-half of each cam chart represents the same amount of time in 
each of the chart drawings. This is because the data are the same 
in all the cams represented in Figs. 71 to 119. To find other points 
on the time-velocity diagram, divide the time-distance curve by a 
number of equally spaced ordinates as shown at J, I, H, Fig. 98. 
The tangent to the curve at K, on the ordinate J V, is K M, and the 
time unit K L is equal to D E. Then, from the same reason- 
ing as given above for the point E, L M becomes a measure 
of the velocity of the follower at K, and it is laid off at M L in Fig. 
100. Similar constructions are repeated at the other points and the 
time-velocity diagram completed. 

270. The time acceleration diagrams are found graphically 
from the time-velocity diagrams by similar constructions. In Fig. 
100 a tangent E S is drawn to the time-velocity curve at E and if the 




~ M 

Fig. 100. — (Enlarged) Velocity Diagram for Circular Base Curve Cam 

velocity of the follower is continued along this line for a time repre- 
sented by E Q it will lose a velocity of Q S in the time E Q. Such loss 
in velocity is retardation and consequently the distance S Q is laid 
off at E D at the center of the time-acceleration diagram in Fig. 101. 
The line E S in Fig. 100 was drawn to the left, and consequently 
downward to make the drawing more compact. In this way retarda- 
tion instead of acceleration was found logically. Had the tangent 
line E S been drawn to the right, and consequently upward, the 
value Q S would have been found just the same and would have been 
called acceleration. The length of the acceleration diagram, A C, 
in Fig. 101 may be taken any value; also, the time unit E Q in Fig. 
100 may be taken any value entirely independent of the time unit 
used in Fig. 98, so long as the same length of line is taken in all the 



140 



CAMS 



velocity diagrams as the time unit, in making comparisons. If a 
definite speed is assigned to the cam then all the lines in the time- 
distance, time-velocity and time-acceleration diagrams will have a 
definite value in feet and in seconds, and by closely following these 
values, the diagrams may be scaled so as to interpret them in the 
ordinary units of feet and seconds, even if arbitrary time lines have 




Fig. 101. — (Enlarged) Acceleration Diagram for Circular Base Curve Cam 

been used in constructing the diagrams. For example, suppose that 
the cam in Fig. 99 is turning at 120 revolutions per minute. Then it 
will require V12 second to turn through the 60° angle DOC, and D E 
in Fig. 98 will represent V24 second. D P measures .5625 inch or 
.0469 foot. Therefore the velocity of the follower at E will be 




ten |Ctt 
Fig. 99. — (Duplicate) Circular Base Curve Cam, Case 1 

.0469 foot per V24 second, or 1.125 feet per second. The scale on 
X E, Fig. 100, would then be graded so that a mark at 1.125 
would fall at E. 

In Fig. 100, A C represents V12 second, and Q E, 1 / 4 s second. 
Since X E represents 1.125 feet per second in this example, Q S repre- 



CAM CHARACTERISTICS 141 

sents .750 foot per second to the same scale. Therefore the accelera- 
tion is .750 foot per second per l ,. s second or 36.00 feel per second 
per second. The scale on ED, Fig. 101, would then be graded so 
that a mark at 36.00 would fall at D. 

Another set of construction lines for obtaining an ordinate in the 
acceleration diagram is shown at L T V, Fig. 100, where L T is the 
same length as E Q, and V T is equal to the acceleration and is laid 
off at V T in the acceleration diagram, Fig. 101. 

271. Degree of precision obtained by graphical method. 
In Fig. 98 the tangent lines may be drawn with precision because the 
curve A E is an arc of a circle, but in the other curves the center of 
curvature for each of the construction points is not known and the 
tangent must, therefore, be drawn by eye. Even here considerable 
precision may be obtained if, in so drawing the tangent, it is remem- 
bered that the tangent at L, Fig. 100, for example, will be practically 
the same distance from U as it is from E when it passes each of these 
points, provided U and E are on ordinates equally spaced, and pro- 
vided also that the curve A E has a fairly uniform rate of curvature 
on both sides of L. If the radius of curvature to the right of L should 
grow noticeably shorter than the radius of curvature to the left of L, 
the tangent at L would pass a little closer to U than to E. If, in 
addition to using such judgment as here indicated in the drawing of 
tangents to irregular curves, a sufficient number of points are taken 
closely together, and if the newly derived curve is drawn smoothly 
through the average positions of plotted points, a remarkable degree 
of accuracy may be obtained by the graphical method of obtaining 
velocity and acceleration diagrams. 

272. Comparison of relative velocities and forces produced 
by cams having different base curves. This comparison, which 
may be made by studying the several velocity and acceleration 
diagrams in Figs. 72 to 121, is also shown in the accompanying table 
where the maximum velocities of the follower are shown in Column 2, 
and the maximum acceleration and retardation values in Columns 3 
and 4. Since force equals acceleration multiplied by mass, the 
direct effort required to move the follower is proportional to the 
acceleration, and, therefore, the relative direct force needed to 
operate the follower for various cams is also shown in Columns 3 and 4. 
The retardation values in Column 4 represent the relative pressures 
exerted by the follower against the cam surface in slowing up where a 
positive drive cam is considered. They also represent the relative 



142 



CAMS 



sizes of counterweights where a gravity return is used. In the cam 
with the straight-line base there would be violent shock at the start 
and the cam would " stick " and require considerable direct power, 
but after that it would be necessary only to overcome friction. The 
parabola, it will be noted from the table and from Fig. 93, requires 
the least direct effort, considering the entire cycle of the follower. 
This effort is represented by unity for purpose of comparison. The 
circular base curve cam, Case II, Fig. 113, requires a trifle less 
effort than the parabola cam while on acceleration on the forward 
stroke, but 2.86 times the effort of the parabola while the follower is 
on acceleration during the return stroke where a double-acting cam is 
used. For a single-acting cam the values given in Column 4 show 
the relative forces necessary to sufficiently accelerate the follower on 
the return stroke so as to keep it in contact with the cam. 



TABLE SHOWING RELATIVE MAXIMUM VELOCITY, ACCELERA- 
TION AND POWER FOR EACH TYPE OF CAM 



Form of Cam 



Col. 1 



All-logarithmic 

Logarithmic combination 

Straight line 

Straight-line combination curve (r = h) 

Crank curve 

Parabola 

Tangential curve, Case I 

Circular curve, Case I 

Elliptical curve 

Cube curve, Case I 

Circular curve, case II 

Cube curve, Case II 

Tangential curve, Case II 



Relative Amounts of Direct 

Force Needed to Operate 

Cam During 




273. Cam follower returned by springs. Although the cam 
built from the parabola chart pitch curve gives the smoothest motion 
and requires the least direct power to operate it so far as the cam and 
follower only are concerned, there may be other considerations in the 



CAM CHARACTERISTICS 143 

design that make or appear to make some other form of chart pitch 
curve more desirable. For example, when a follower is returned by a 
positive drive parabola cam, or when it is returned by gravity, the 
parabola cam gives the best action because the pull on the follower 
is constant all the time, but when the follower is returned by a spring, 
the spring reacts on the cam with a uniformly increasing pressure 
during the out stroke as represented by the straight inclined clash- 
line & P in Fig. 93, and with a reverse uniformly decreasing pressure 
during the instroke. 

274. If the spring pressure acting on the cam is zero when 
the follower is at rest in its lowest position, the spring compression 
line would be represented by the straight line A N, Fig. 93, starting at 
A and inclined so as to touch the retardation 
line as at F. Inasmuch as there should 



jp 



always be some compression in the spring, s ^'^•■y 
even when the follower is at rest, a margin ^r 

of compression will be taken as illustrated 
at A S. The practical spring compression Fig. 93.— (Duplicate) Accel- 
line will, therefore, be S P parallel to A N. pakabola Cam**' 
As the follower moves out, its acceleration 

during the first part of the stroke produces increasing pressure between 
the cam surface and the spring-actuated follower as represented by 
the increasing length of the ordinates from S B to R D. At mid- 
stroke the follower begins to slow up. In the case shown in Fig. 93, 
the slope of the spring pressure line was taken so as to have the same 
spring pressure (R F = S A) on the cam at midstroke as it has at the 
beginning. The line S P could have been given a steeper slope if a 
larger margin of pressure than R F had been desired at midstroke. 
This would have required a heavier spring. From midstroke to the 
end there is again an increasing margin of pressure, the maximum 
being represented by the difference between the ordinates P H and 
R F. The full strength of the spring which would have to be used 
would be represented by the ordinate P C. 

275. Relative strength of spring required for crank, tan- 
gential, CUBE AND PARABOLA BASE CURVE CAMS. Although the 

parabola cam, with its perfect action as described in paragraph 214, 
permits of the use of a light spring when a single spring is used to 
return the follower, the crank curve, tangential curve and cube 
curve cams may each be designed to operate with somewhat lighter 
springs. Spring compression lines for each of the three last-men- 



144 



CAMS 



tioned cams are shown at S P in Figs. 89, 97, and 109, and the max- 
imum compression required of a single spring in each case is 1.75, 
2.35, and 2.30 as compared with 2.40 for the parabola cam as shown 
in Fig. 93. The return spring pressure between the follower roller 
and the cam surface, when the crank base curve is used, is more 
nearly uniform throughout the entire stroke than it is with any other 










D 


: 2 
1 


A. 






E 




S L 


-^ 


^--^ 


F 


— 1 




Fig. 


109. 


~R 

Q 


5^2^*^- 



-2 RP 

Fig. 89. — (Duplicate) Acceleration Diagram for Crank Curve Cam 

Fig. 97. — (Duplicate) Acceleration Diagram for Tangential Cam 

Fig. 109. — (Duplicate) Acceleration Diagram for Cube Cam 



type of cam, as may be noted from the maximum and the average 
ordinates between the acceleration-retardation curve and the spring 
pressure line, S P, in the several diagrams. 

276. Cube curve cam specially adapted for a follower returned 
by a spring. The cube curve cam possesses one characteristic over 
the others in that the pressure between the cam and the follower is 
absolutely uniform during the latter part of the up-stroke and the 
first part of the down-stroke when the follower is returned by a 
rpring, as shown by the parallel lines F H and R P, Fig. 109. This 
gives an advantage of smooth running and uniform wear when the 
spring is under its greatest compression. 

277. The pressure between the spring-actuated follower 
and the cam is variable throughout the stroke in all cams except 
during part of the stroke with the cube curve cam. And it may 

readily happen that the acceleration called 
for by the cam is so great that the spring 
will not be strong enough to keep the fol- 
lower roller against the cam surface as may 
be specially noted at or near the beginning 
of the return stroke. This is illustrated in 
(Duplicate) Accel- Fig. 113 where the spring pressure against 
eration diagram for Cir- | ne follower which would be necessary to 

cular Base Curve Cam, . ,11 n tti 

Case2 hold it to the cam is represented by t L, 

whereas, if a spring of the same strength as 

for the cube curve cam, Fig. 109, were used the pressure at the 




CAM CHARACTERISTICS 



145 



phase E, Fig. 113, would be only R E. This means that the cam 
will "run away" from the follower, because the spring is not strong 
enough during the part of the stroke represented by TF R to 
press the follower against the rapidly receding cam surface. 

278. In order to keep the follower roller against the 
cam surface where cams with large retardation values are used, as 
in Figs. 77, 85, 113, 117, and 121, a comparatively heavy spring is 
required which will be unnecessarily strong during a very large part 
of the stroke, or else two springs will be required, the second one to 
come into action when needed. Both cases are illustrated in Fig. 113. 
A single heavy spring that will exert a pressure represented by W V 



:nr 


r 2 
1 




Fig. 77? 


r-1 


z 




Fig. 85.br 




Fig. 77. — (Duplicate) Acceleration Diagram for Logarithmic-Combination Cam 

Fig. 85. — (Duplicate) Acceleration Diagram for Straight-Line-Combination Cam 

Fig. 117. — (Duplicate) Acceleration Diagram for Cube Cam, Case 2 

Fig. 121. — (Duplicate) Acceleration Diagram for Tangential Cam, Case 2 

will keep the follower roller against the cam surface at all times, the 
minimum pressure between the two occurring at F G. Or, a single 
and much lighter spring exerting a pressure represented by S P, Fig. 
113, may be used, and then a second and shorter spring with an initial 
compression represented by M E may be so placed as to come into 
action at E so that the combined pressure of the two springs on the 
follower is M E plus R E equal F E. This means that the combined 
pressure of the two springs will be just sufficient to keep the follower 
roller against the cam at phase E, and that the total pressure of the 
two springs at the end of the stroke will be represented by C V, thus 
giving an excess pressure represented by H V at the end of the stroke. 



146 



CAMS 



The base curves that are best suited for spring-return followers will 
be seen by comparing the acceleration diagrams in Figs. 73 to 121 
to be the crank curve, parabola, tangential curve, Case I, and the cube 
curve, Case I. The logarithmic combination and straight-line com- 
bination curves come next in order. 

279. Accuracy in cam construction. It need scarcely be 
pointed out that the pitch surfaces of cams should be constructed 
with considerable accuracy and the working surfaces carefully fin- 




15 r 



Fig. 111. — (Duplicate) Circular Base Curve Cam, Case 2 

ished, if definite results are required, for, it may be seen by com- 
paring the pitch surfaces of several of the cams illustrated in Figs. 
71 to 119 that a relatively small difference in form may make a large 
difference in the velocity, acceleration, and force or pressure, under 
which the follower operates. For example, the circular curve cam, 




Fig. 103. 
Fig. 107.- 




-(Duplicate) Elliptical Base Curve Cam 
-(Duplicate) Cube Base Curve Cam, Case 1 



Case II, Fig. Ill, and the cube curve cam, Case II, Fig. 115, are 
apparently quite similar in form, though varying in sizes, yet the 
maximum accelerations which they impose on the follower on the 
return stroke are quite different, being 2.9 and 4.8 respectively, as 
shown in Figs. 113 and 117. Also the cube curve cam, Case I, Fig. 
107 and the elliptical cam, Fig. 103, are much alike, yet their velocity 



CAM CHARACTERISTICS 



147 



lines and their acceleration lines, Figs. 109 and 105, arc different in 
every way and if a spring were used to return the follower, the one 
for the elliptical cam would have to be enough heavier to carry 1.7 
more compression at the end of the stroke than the one for the cube 
cam, assuming an initial pressure of A S, in each one. The value 
1.7 is found by comparing the lengths C P in Figs. 109 and 105. 





FlG. 105T 



Fig. 105. — (Duplicate) Acceleration Diagram for Elliptical Base Curve Cam 
Fig. 109. — (Duplicate) Acceleration Diagram for Cube Cam 

280. Regulation of noise. If a cam follower, as for example a 
cam-operated disk valve, comes to rest on a seat at one end of its 
stroke, it is evident that it would be desirable for the follower to have 
the least possible velocity for at least a short distance before it reaches 
the seat, in order to provide against unnecessary striking velocity. 
Noise will be in some proportion to the velocity of the follower at 
the instant of seating. With this in mind, an examination of the 
velocity diagrams in Figs. 72 to 120 will show that the cube base 
curve, Case I,' Fig. 106, gives by far the best results, for, the vertical 
ordinates of the velocity curve in Fig. 108 are very much smaller 
as the follower approaches A than they are in any other diagram, 
excepting Case II of the cube curve, Fig. 116, but in this instance the 
advantage is more than offset by the high retardation values at the 
end of the stroke as shown in Fig. 117. The circular curve, Case II, 
comes next in the matter of giving small velocity to the follower, 
Fig. 112, but it does not possess the advantage of the cube curve 
when a spring is used to return the follower. The crank curve cam is 
least adapted of all the cams where quiet seating of a follower is 
desired, as may be observed by noting that the velocity curve, 
Fig. 88, for this cam is convex upwards, whereas the others are 
straight or convex downwards and thus have smaller initial vertical 
ordinates and, therefore, lower velocity. The full practical ad- 
vantage of cams which give low-sealing velocities and consequently a 
more quiet follower action, is offset to a considerable extent where 
the follower operates a valve which must admit a comparatively 
large volume of gas or fluid quickly. 



148 CAMS 

281. High speed cams. Cams intended for use on high-speed 
machines should give the smoothest possible motion to the follower, 
that is, should be free from sudden variations of velocity during the 
stroke and from shock due to sudden starting and stopping. A 
study of the velocity diagrams, Figs. 72 to 120, shows that the all- 
logarithmic and the straight-line base curves, Figs. 72 and 80, give 
extreme velocity right at the start in all cases; and that the logarith- 
mic-combination and straight-line combination cams will also give 
relatively high velocities at the start, Figs. 76 and 84. Therefore 
none of these cams would, in general, be suitable for high-speed work. 
Among the other cams some have an advantage at one end of the 
follower stroke where the rate of change in velocity is low, but they 
lose it at the other end where it is high as, for example, the cube cam 
Case II, as shown in Fig. 117; or they lose their advantage at the 
center or some intermediate point as in the elliptical cam, Fig. 105. 

282. The cams specified in the preceding paragraph give rela- 
tively large sudden change of velocity to the follower either at one 
end of the stroke or the other, or at intermediate positions; and of 
the remaining cams, the parabola cam is the only one that gives 
absolutely uniform rate of change of velocity to the follower. The 
crank curve, the circular curve, Case I, and the tangential curve, 
Case I, give relatively good results, all being at a slight disadvantage 
compared with the parabola due to variations in acceleration of the 
follower. . This disadvantage, however, is small, and these three 
cams, together with the parabola cam, should give best results 
where there is high speed, provided they are accurately designed and 
made. 

283. Balancing of cams. In addition to the forms of the curves 
here discussed for the pitch surfaces of cams that are to run at high 
speed, it is necessary to design the cam and so place the weight that 
the cam will be as nearly balanced as possible. This matter of bal- 
ancing is one of the greatest drawbacks to the use of the cam in high- 
speed work, for the very nature of a cam implies irregularity in form 
and hence difficulty in balancing. The face cam cut on a full cir- 
cular disk as illustrated in Fig. 2 comes nearest to a natural balance 
of any of the forms of radial cams. The trouble due to lack of 
natural balance in ordinary radial cams may easily be so decided as to 
render them quite impracticable in many cases where high speed 
and large stroke are required, unless elaborate balancing problems 
are solved in connection with the cam design. Small radial cams 
with small strokes have been made to run at exceedingly high speeds. 



CAM CHARACTERISTICS 



149 



The cylindrical cam, because of its natural balanced form with respect 
to the axis of rotation, is we'll adapted to high speeds. 

281. Pressure angle factors for 20°, 30°, 40°, 50°, and (i()° 
for various FORMS of CAMS. Most of the base curves for cams are 
of such nature that it is only necessary to multiply the follower 
motion by a given factor and then multiply the product by 3(30 and 
divide by the number of degrees the cam rotates during the follower 
motion, to obtain the circumference of the pitch circle and the proper 
size of the cam for a given pressure angle. The logarithmic and tan- 
gential base curves are of such a nature that no one factor can be 
used for all data that include a common pressure angle. When these 
base curves are used the length of chart, if desired, must be com- 
puted by separate formulas for each problem. The logarithmic and 
tangential base curves are most easily applied by constructing the 
cam pitch surface directly from calculated values in each problem 
without the use of any chart whatever. 

285. The factors for pressure angles for all base curves, excepting 
the logarithmic and tangential, are given in the accompanying 
Table of Factors for 20°, 30°, 40°, 50° and 60°. These factors are 
also laid off graphically in Fig. 132, thus enabling one to use inter- 
mediate values if desired. For partial comparison of the curves 
which have no general factor with those which have, the special fac- 
tor in each case for the single comparative problem which has been 
used throughout in designing the cams in Figs. 70 to 121 is given in 
the following paragraphs, and these factors are plotted to give the 
dash lines in the accompanying chart for factors. 

286. Varied forms of fundamental base curves. Several of 
the base curves are, or may be, used in practical work with variations 



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in details of construction, as, for example, in the straight-line com- 
bination curve, Fig. 82, the easing-off arc A E has a radius A B equal 
to the total rise of the follower, whereas it would be equally correct 



150 



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152 



CAMS 



in principle to make this radius Y^AB. In this latter case the cam 
would be smaller for a given pressure angle, but the shock on starting 
and stopping would be greater. This case is not illustrated in Figs. 
70 to 121 but is included under item 4 in the Table of Pressure Angle 
Factors, and also in the Chart of Pressure Curves, Fig. 132. Like- 
wise the factors for the elliptical base curve having a ratio of 2 to 4 
instead of 7 to 4, are given in item 5 in the Table and also in the 
Chart, Fig. 132. The factors for a cube base curve made up of sym- 
metrical cube curves are also given in item 13 in the Table where it 
may be noted that this base curve gives an extremely large cam 
where small pressure angles are desired. 

287. Methods of determining the cam factors. The methods 
of computing the cam factors for various base curves are briefly 
described in the following paragraphs. The letter h in the following 
formulas represents the motion of the follower, and the letter a the 
maximum pressure angle. 

288. All-logarithmic and logarithmic-combination curves. 
These base curves do not have a constant factor for each pressure 




Fig. 78. — (Duplicate) Straight Base Line 

angle. The radius for the pitch circle in each problem is found by 
computation and graphics as described in paragraph 182 et seq. 

The factors for the data used in the charts shown in Figs. 70 
and 74 are: 

For all-logarithmic cam: 20°, 2.28; 30°, 1.28; 40°, .76; 50°, .42; 
60°, .21. 

For logarithmic-combination cam: 20°, 2.76; 30°, 1.69; 40°, 
1.04; 50°, .62; 60°, .34. 

?89. Straight-line base. Fig. 78. 



AR = FC cot a = hX cot a = IX 1.73 = 1.73. 



CAM CHARACTERISTICS 



153 



290. Straight-line combination base curve, Fig. 82. 

AR = 2AN + 2N X = 2h tan ( | j -f kota = 2 X 1 X .268 

+ 1 X 1.73 = 2.27. 

291. Crank curve, Fig. 86. This curve may be regarded as the 
projection of a helix and, therefore, D Q equals the length of the 




Fig. 86. — (Duplicate) Crank Base Curve 

quadrant R which in turn is equal to^irh. The line E Q is tangent 
to the base curve at E. 

AR = 2DE = 2DQXcota = 1.57 h cot a = 1.57 X 1 X 1.73 = 
2.72. 

292. Parabola, Fig. 90. In a parabola, the subtangent D Q 
is equal to twice the projected length of the curve A E, and, therefore, 
DQ = h 

AR = 2 DE = 2 DQ cot a = 2 h cot a = 2X 1 X 1.73 = 3.46. 

c 




Fig. 90. — (.Duplicate) Parabola Base Curve 

293. Tangential curve, case I, Fig. 94. This curve has no 
common factor for a given pressure angle and the radius of its pitch 




Fig. 94. — (Duplicate) Tangential Base Curve, Case 2 

surface must be computed directly by formulas given in paragraph 
222 without the intervention of a cam chart. For purposes of com- 
parison with other curves the following factors are given ; they apply 



154 



CAMS 



only for the data that have been used in the cams illustrated in Figs. 
71 to 119. 

20°, 5.28; 30°, 3.62; 40°, 2.82; 50° 2.36; 60°, 2.09. 

These values are shown in the dash line curve, No. 9, in Fig. 132. 
294. Circular curve, case I, Fig. 98. The chord E C is per- 
pendicular to the line S T which bisects the angle C B E. This angle 



f Y \ 




Fig. 98. — (Duplicate) Circular Base Curve, Case 1 

is equal to the pressure angle. The line E F is perpendicular to C S. 
Therefore angle C E F equals one-half of the pressure angle. Then 

E F = F C cot Y 2 a and 

A R = 2 E F = h cot Y 2 a = 1 X 3.73 = 3.73. 

295. Elliptical curve, Fig. 102. The length of the cam chart 
for the elliptical curve for a pressure angle of say 30° may be most 



Tti Tve C L y 




Fig. 102. — (Duplicate) Elliptical Base Curve 

readily found by constructing several arbitrary elliptical charts, say 
four, each with a pressure angle factor, or length, of 2, 3, 4, and 5 re- 
spectively and each having a common height equal to the rise of the 
follower. Having constructed the elliptical curve in each of the charts, 
draw tangents in each case as at E, Fig. 102, and measure the angle 
E N X which will be the pressure angle corresponding to the factor 
or length assumed. Then, on any coordinate paper plot a curve 
with the pressure-angle factors as ordinates and the corresponding 
measured angles as abscissas. This curve will cross the ordinate 



CAM CHARACTERISTICS 



155 



which passes through the assigned pressure angle, in this case 30°, 
and the length of ordinate will give the desired cam factor. 

296. Cube curve, case I, Fig. 106. The pressure angle factors 
for this case in which two unsymmetrical cube curve arcs are used 




Fig. 106. — (Duplicate) Cube Base Curve, Case 1 

are specially computed by the formulas given in paragraph 242. 
The value of I in formula (1) when h = 1, will give the factor for 
whatever pressure angle is assigned to a. For a pressure angle of 30° 

I = 2.427 h cot a = 2.427 X 1 X 1.73 = 4.20. 

297. Circular base curve, case II, Fig. 110. The complete 
factors for this curve are the same as for the circular base curve, 




Fig. 110. — (Duplicate) Circular Base Curve, Case 2 



Case I, and are found in the same general way. In Case I the two arcs 
making up the base curve are equal; in the present case, they are 
unequal, and the formula deduced in paragraph 294 must be used for 
each arc. In this case, the first circular arc is required to lift the 
follower during % of its stroke, and, therefore, the distance A X 
in Fig. 110 will be, 

A X = .75 h cot Y 2 a = .75 X 1 X 3.73 = 2.80. 

The second circular arc is used for the balance of the stroke and, 

therefore, the distance, X R = .25 h cot Y 2 a = .25 X 1 X 3.73 = .93. 

298. Cube curve, case II, Fig. 114. In this case the cube curve 

is used for % of the stroke and a circular arc for the remainder of the 



156 



CAMS 



stroke. The formula x = is used to compute the part ilof 

Tan a 

the cam chart length. The value of h is the follower's total motion 




Fig. 114. — (Duplicate) Cube Base Curve, Case 2 

and that of / is the fractional part of the follower's motion during 
which acceleration takes place. Then 

AX = 3X 'J!L X 1 = 3.90. 

.577 

The length X R is found in the same manner as in the preceding 
paragraph and is the same value, namely .93. 

299. Tangential curve, case II, Fig. 118. This curve, like 
Case I of the tangential cam has no cam chart, unless it is specially 




Fig. 118. — (Duplicate) Tangential Base Curve, Case 2 



desired to lay it out after the cam is drawn by making special com- 
putations based on the pitch circle as described in paragraph 225. 
For purposes of comparison the data used in this cam, as drawn in 
Fig. 119, are the same as for all other cams in Figs. 71 to 119, and for 
the data so used the pressure angle factors are: 

20°, 13.02; 30°, 5.86; 40°, 3.36; 50°, 2.20; 60°, 1.57. 

These values are shown in the dash line curve, 16, in Fig. 132. 



SECTION VIII.— MISCELLANEOUS CAM ACTIONS AND 
CONSTRUCTIONS 

300. Variable angular velocity in the driving cam shaft. 
The subject of variable angularity velocity in the drive shaft of a cam 
applies to all types of cams, but it is rarely met with except in oscil- 
lating cams. The reason for this is that in machinery, in general, 
the shafts that make a full turn do so with practically uniangular 
velocity except in slow-advance and quick-return motions and 
in some special cases, and, therefore, the shaft that operates a cam, 
in general, is considered to have uniform angular velocity. But 
with the oscillating cam the motion must come through a crank and 
connecting rod, or eccentric and beam, or some other device, from a 
shaft which, in general, turns with uniform angular velocity, and 
which gives to the oscillating cam a variable angular velocity as 
illustrated in Fig. 133 where the unequal arcs B\ G\, G\K\, K\L\ 
represent the distances traversed by the cam pin B\ while the main- 
shaft crank pin turns through the equal arcs B G,G K and K L. The 
method of building a cam which has variable angular velocity will be 
illustrated in the following problem. 

301. Problem 29. Oscillating cam having variable angular 
velocity, toe and wiper type. Required an oscillating wiper cam, 
operated by a crank and connecting rod from a n ain shaft to raise 
and lower a straight-toe follower through a distance of one unit while 
the crank shaft turns through 120°. Assume the following dimensions : 
Main crank radius, C B, 4 units, Fig. 133; connecting rod length, 
B B\, 20 units; cam-arm radius, B\ 0, 5 units; shortest cam surface 
radius, A, 2 units. Find the distance the follower will move 
during each of three equal periods of time on the up-stroke. 

302. The first step in the solution of the problem is to lay out the 
main crank center as at C in Fig. 133; then the crank-pin circle with 
a radius C B of 4 units, and next the connecting rod length of 20 
units on the centerline as at E J. Lay off the assigned 120° of crank- 
shaft motion symmetrically about the main centerline as at BCD 
and with B and D as centers and the length of the connecting rod 

157 



158 



CAMS 



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MISCELLANEOUS CAM ACTIONS AND CONSTRUCTIONS 159 

as a radius draw two arcs intersecting on the horizontal centerline, 
thus locating B\. With C as a center and the connecting-rod plus 
the crank as a radius, draw the arc passing through «/; with C as a 
center and the connecting rod minus the crank as a radius, draw the 
arc passing through J] . 

303. To find the center of the cam shaft, Fig. 133, take Bi as 
a center and the assigned cam-arm radius of 5 units, and draw an 
arc, on which the point will be found later. On this arc find a 
point, by trial and error with the compass, which is the center of an 
arc which passes through B\ and which intersects the two arcs through 
J and Ji at the same elevation, as, for example, at L\ and l\. The 
center point so found is the point 0. The arc L\B\Fi will then be 
the arc of swing for the center of the cam-arm pin, and the angularity 
of action between the connecting rod and the cam arm at the two 
extreme ends of the cam-arm swing will be approximately the same. 
Draw a vertical line through and mark the assigned distance A 
which is the shortest radius of the cam surface. The horizontal line 
through A will be the lowest position of the flat-surface follower toe. 
The distance A V is equal to the assigned motion for the follower. 

304. Having completed the general layout of the assigned data, 
the cam surface A V2 is found as follows: Draw the arc Bo Lo with 
a radius equal to Bi, and make the length B2 L2 equal to B\ L\. 
Revolve V about until it meets the radial line drawn from L2 to 0, 
thus determining the point V\. At this latter point draw a line 
Vi V2 perpendicular to V\. With the aid of any smooth-edged 
curved ruler draw a curved line tangent to A W at A and also tan- 
gent to Vi V2 at the point where it happens to come. Such a curved 
line is shown at A V2 in Fig. 133. Any other curved line tangent 
to the straight lines A W and V\ V2 would have done the work 
in the same time but would have given slightly different intermediate 
velocities to the follower as will be explained in a later paragraph. 

The actual working length A W of the follower toe is readily 
obtained by revolving the point of tangency V2 about until it 
meets the horizontal line through V at Vs. Projecting V3 down to 
A W and adding a short distance W W\ to prevent a sharp-edge 
action, the practical length A W\ is obtained. If the toe shaft is 
offset a distance A Y the total length of follower toe will be Y W\. 

305. To find the distances moved by the follower toe during each 
of three equal periods while on the upstroke, divide B L, Fig. 133, 
into three equal parts as at G and K. With these points as centers 



160 CAMS 

and with the connecting rod length as a radius construct short arcs 
intersecting B\ L\ as at Gi and K\ . Lay off the arcs Bi G\ and B\ K\ 
at B2 G2 and B2 K2 and draw the radial lines G2 and K2. Per- 
pendicular to these radial lines draw other straight lines, tangent 
to the curved cam surface A V2, thus obtaining the lines Hi H2 and 
1 1 1 2. Revolving H\ and I\ back to the vertical line, the points H 
and / will be obtained and the distances moved by the follower during 
the three equal time periods on the upstroke will be A H, H I and 
I V respectively. 

306. The path of contact between the cam wiper and the toe is 
shown by the curved dash line A Vs, Fig. 133. Points on this curve, 
such as at Is, are obtained by revolving the point of tangency 1 2 
around until it meets the horizontal line through /. 

307. Other considerations relating to variable angular 
velocity drive, brought out in this problem (Problem No. 29) are 
that the follower toe takes a longer time for the down-stroke as shown 
by the length of arc L D as compared with L B, Fig. 133. This could 
be rectified and both times made the same, if desired, by placing the 
center of the cam so that the points B\ and L\ would be on the 
horizontal line through C. This would only be possible with certain 
limited combinations of lengths of crank arms and rods, and in any 
event the intermediate velocities of the follower would be different 
on the up- and down-strokes. If it were desired to know the distances 
moved by the follower during three equal periods on the down- 
stroke the equally spaced points M and N, Fig. 133, would be obtained 
and used in exactly the same way as explained for G and K in para- 
graph 305. 

The point F is the outward dead center position of the driving 
crank pin and is found by continuing the straight line through F\ 
and C to F. When the driving crank pin is at F, the cam surface 
is in the position shown by the dash line A3 W5 and A\ is at A. 
While B is moving from D to F, A\ is moving to A and the follower 
toe is at rest, being supported by the cylindrical surface A A\ rub- 
bing against it, or it may be supported by a resting block indicated 
at £7. 

It is sometimes thought that this toe-and-wiper cam is prac- 
tically free from rubbing action especially where the length of the toe 
surface equals approximately that of the wiper, but it will be seen 
from the velocity diagram shown just above the cam and described 
in paragraphs 317 and 318, that there may be considerable rubbing. 



MISCELLANEOUS CAM ACTIONS AND CONSTRUCTIONS 1G1 



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162 CAMS 

There must be some sliding in all flat-toe followers where the acting 
surface is perpendicular to the right-line motion of the follower, as 
it is in Fig. 133. 

308. Exercise problem 29a. Oscillating cam having vari- 
able angular velocity. Required an oscillating wiper cam, oper- 
ated by a crank and connecting-rod from a main shaft, to raise and 
lower a straight-toe follower through a distance of three units while 
the crank shaft turns through 150°. Find, also, the distances that 
will be traversed by the follower toe during equal intervals of time on 
the up-stroke. Assume the following dimensions: Main crank 
radius, 5 units; connecting-rod length, 30 units; cam-arm radius, 7 
units; shortest cam surface radius, 4 units. 

309. TOE-AND-WIPER CAM WHERE TOE IS CURVED. In the toe- 

and-wiper cam explained in the paragraphs immediately preceding, 
a flat surface toe Y W, Fig. 133 was used. A curved toe such as is 
shown at A W, Fig. 134 may be used as illustrated in the following 
problem. 

310. Problem 30. Required a wiper cam to operate a 
curved-toe follower which shall move: 

(a) Up 4 units on the elliptical base curve where the ratio of 
axes is 2 to 4, while the cam turns 45° in a counter-clockwise direc- 
tion with uniform angular velocity. 

(b) Down 4 units on the same base curve, while the cam turns 45° 
in a clockwise direction with uniform angular velocity. 

311. While the follower toe may have the form of any smooth 
curve which is convex to the cam wiper, an arc of a circle will be 
assumed because of the ease in drawing. The general principles of 
construction are the same for this problem as in Problem 12. The 
shortest radius A of the wiper cam, Fig. 134 is assumed. The 
form of the curved toe is the circular arc A W, with its center at Ai. 
It is convenient in such a problem as this to work with the center 
points of the follower arc, and, therefore, the 4 units of travel are laid 
off first at A\ Vi instead of A V. The semi-ellipse in which h Ui: 
IiVi : : 2 : 4 is drawn and the perimeter divided into equal parts 
at J', Ui, H' . Only four construction points are used in this problem 
in order to secure as much simplicity as possible in the illustration. 
In practice, more construction points should be used. The four 
construction centers at Hi, h, Ji, Vi, are revolved to their corre- 
sponding positions relatively to the cam at H2, 1 2, J 2, and V2 and 



MISCELLANEOUS CAM ACTIONS AND CONSTRUCTIONS 163 

B _ 

Fig. 137. Diagram 
or Sl'dino Velocixt 




a* v< ./, H t 




Fig. I'M. — Pkublem 30. Oscillating Cam with Ctjkved-Toe Follower 



164 CAMS 

the toe-arcs drawn as shown at #3, h, Js, and V3. The wiper cam 
curve A C is then drawn tangent to these arcs and the tangent points 
revolved back to their actual positions at Hi, h, J a, and F4, thus 
obtaining the locus of contact between the wiper and toe. This 
locus is shown by the dashline curve A H4 F 4 . The necessary length 
V Y\ of the follower arc is also obtained by projecting the extreme 
point Y on the locus to Y\ and adding an arbitrary distance such as 
Y\ W\ to avoid wear at the tip end. 

312. If an irregular curve had been used for the form of the toe 
instead of a circular arc it would have been necessary to construct 
a template of the desired form of the toe and to move it out radially 
the desired distances on each of the radial construction lines H2, 
1 2 • . ., keeping the template always in the same relative position 
with each of the radial lines. At each of the four adjustments of the 
template, arcs would have been drawn against the template edge 
and the work then continued as described in the preceding para- 
graph. 

313. The pressure angles in the toe-and-wiper cams are quite 
different for flat and curved toes. In Fig. 133 the line of pressure 
is always parallel to the axis Y Y\, of the follower rod, as illustrated 
by the vertical line at W Vs ; and the maximum leverage with which 
it acts on the bearings is Y W. With the curved-toe wiper the line 
of pressure is an inclined line and the pressure angle at the top of 
the stroke is V V\ V±, Fig. 134, and when the follower is half way up 
the pressure angle is 1 1\ I±. 

314. Exercise problem 30a. Required a wiper cam to oper- 
ate a curved-toe follower which shall move: 

(a) Up 3 units with uniform velocity while the cam turns 60° in a 
counter-clockwise direction with uniform angular velocity. 

(b) Down 3 units with uniform velocity while the cam turns 60° 
in a clockwise direction with uniform angular velocity. 

315. Rate of sliding of cams on follower surface. The 
rubbing velocity of cams which are in sliding contact with the fol- 
lower, may be readily determined by constructing simple velocity 
diagrams at each of the construction points, as explained in the fol- 
lowing paragraphs. 

316. Problem 31. Rate of sliding between cam and flat 
follower surfaces. Find the curve of rubbing velocity between 
surfaces in a toe-and-wiper cam mechanism where the follower toe is 



MISCELLANEOUS CAM ACTIONS AND CONSTRUCTIONS 165 

a flat surface. Assume that the wiper oscillates with uniform angular 
velocity. 

317. In Fig. 135 let the angle I± h represenl the uniform angular 

velocity of the wiper cam. Then the point li on the cam will have 
the linear velocity I± 1 5. Laying this value off at h h, where 1 2 

Fig. 136— Diagram of Velocities 




Fig. 135. — Problem 31. — Sliding in Toe-and-Wipeb Cams 



comes into action, and taking the component h It, the actual rubbing 
velocity is obtained. This may be transferred to h h in Fig. 136 
and, finding other ordinates, the complete sliding-velocity curve 
A\ V7 is obtained. The ordinate A A\ is quickly obtained, for it is 
obviously equal to the linear velocity line A A\ in Fig. 135. In 
Fig. 135 the detail construction for obtaining the velocity of the fol- 
lower is shown only at one point, h, the construction at the other 



166 CAMS 

points being the same. Also all lines pertaining to the construction 
of the cam are omitted, as they are fully given in Fig. 52. 

318. The actual rate of sliding in feet per second may be 
readily found at any position by means of the velocity diagram 
in Fig. 136. For example, if the cam shaft 0, Fig. 135, is consid- 
ered to oscillate back and forth through 45°, 100 times per minute 
with uniform angular velocity, and if the radius I± is 14 inches, the 
line I4 1 5 will be drawn to represent a velocity of 

2 X 14 X 3.14 X 2 X 100 . A . . . , 

8 X 12 X 60 " = 3 ' 04 feet Per SeCOncl 

This value, laid off as the resultant velocity at 1%, gives a component 
or sliding velocity I3 I7 which is laid off at 73 I7 in Fig. 136. Other 
ordinates, found in the same way, will give the curve A\ V7, showing 
the sliding velocity between cam and toe in feet per second. The 
minimum rate of sliding will be A A\ shown in both Figs. 135 and 136, 
and will be 1.6 measured on the same scale that was used to lay out 
hh. 

319. The velocity of the follower, in feet per second, may 
also be readily found by simply taking the vertical component ^3 Is, 
Fig. 135, and laying it off at h Is in Fig. 136. Taking the vertical 
components at other points the line A Vs, showing the linear velocity 
of the follower will be obtained. The line A Vs is a straight line in 
this problem because this cam illustration was taken so that the 
follower would have uniformly increasing velocity. In general, 
where the cam curve A V2 in Fig. 135 is assumed, the line A Vs in 
Fig. 136 will not be straight. 

320. Exercise problem 31a. Sliding velocity between cam 
and follower. Assume a flat-toe follower with a rise of 3 inches 
and a cam wiper of minimum radius of 4 inches which oscillates with 
uniform angular velocity through 150 cycles per minute. Construct 
the toe-and-wiper surfaces and find the curve of sliding velocity 
between them in feet per second. Find also the curve of linear veloc- 
ity of the follower toe and state the maximum velocity in feet per 
second. 

321. Problem 32. Sliding velocity with curved toe fol- 
lower. Find the curve of rubbing velocity between surfaces in a 
curved toe-and-wiper cam mechanism, assuming that the wiper 
oscillates with uniform angular velocity. 



MISCELLANEOUS CAM ACTIONS AND CONSTRUCTIONS 167' 

In curved-toe followers the general principle of obtaining the 

rubbing velocity is the same, although the del ail of drawing the 
velocity diagram differs slightly. In Fig. L34 the linear velocity of 
the point Hz on the cam is //.-, He and this value is laid off at II \ II-. 
The direction of sliding at this phase must be that of the common 
tangent line to the two surfaces, and its length, which represents the 
velocity of sliding, is found by drawing the line H7 Hs parallel to the 
direction of motion of the point 1I± on the follower. The length of 
H4 Hg is thus found and is laid out as shown in Fig. 137, directly over 
H± of Fig. 134. Other lines representing the rubbing velocity are 
similarly found and laid out in Fig. 137, thus obtaining the rubbing 
velocity curve AiUs Vg. 

322. In the case of the curved-toe follower it will be noted that 
that portion of the toe from V4 to Yi, Fig. 134, will be traversed 
twice as often as the portion from V to F4, and in addition the rubbing 
velocity will be much greater. In the flat-toe follower, Fig. 135, the 
point of contact travels regularly forth and back the full distance on 
each stroke, but the wear as in the curved-toe follower will be irreg- 
ular, due to the variable rubbing velocity, which in the case illustrated 
in Fig. 136 is a maximum at the tip end. 

323. Exercise problem 32a. Sliding velocity with curved- 
toe follower. Find the curve of rubbing velocity between cam 
surfaces in Problem 30a, assuming that the wiper cam oscillates 
through a cycle 90 times per minute. Show scale for curve. 

324. Problem 33. Sliding velocity where cam has variable 
angular velocity. Find the curve of rubbing velocity between 
surfaces in a flat toe-and-wiper cam construction, assuming that the 
wiper cam oscillates with a variable angular velocity. 

325. When an oscillating cam has variable angular velocity, as 

in Fig. 133, the extent of the sliding action between cam and follower 

may be found as in the present example. In Fig. 133, the length 

of crank represented by C E is 4 inches and the crank is assumed to be 

turning 120 revolutions per minute. The velocity of the crank pin 

... . 2X 3.14 X4 X 120 

will then be r ^ f - = 4.19 leet per second. 

1Z X. OU 

326. The velocity just obtained is represented by the line K U, 
Fig. 133, laid off to any convenient scale. Its component K L\ along 
the rod is found by dropping from U a perpendicular to the con- 
necting-rod position KK\. The component K U\ is then trans- 
ferred to the other end of the rod at K\ Uo- This component gives a 



168 CAMS 

resultant linear velocity of K1U3 to the cam crank pin at the phase K\. 
At the radial distance K3, which is equal to the radii 01 2, and 1 3 
the linear velocity will be K3 U± and this transferred to I3 will give 
I3 U5 as the resultant linear velocity of I2 when it becomes the 
driving point. The line 73 Uq is the component in the direction in 
which sliding must take place and this is laid off at I3 Uq in Fig. 138. 
If K U represents 4.19 feet per second, I3 Uq, will represent 1.30 feet 
per second to the same scale and the maximum velocity of sliding, 
which is represented at A3 Aq, will be 1.87 feet per second. 

327. Exercise problem 33a. Sliding velocity where cam 
has variable angular velocity. Assume crank C E, in Fig. 133, 
to be 5 units long and turning at rate of 100 revolutions per minute; 
also, take the angle B C D = 150° symmetrical about C E, the con- 
necting rod B Bi = 30 units, the cam arm Bi = 7 units, the mini- 
mum cam radius A = 4 units and the cam lift 3 units. Construct 
the cam and follower and draw the curve of sliding velocity to scale. 

328. Elimination of sliding friction where flat or curved 
surface followers are used. The ordinary toe-and-wiper cam 
mechanism operates with more or less sliding action as shown in the 
preceding paragraphs. Cams resembling the toe-and-wiper type 
may be constructed so as to eliminate all sliding friction by using 
special curves and lines for the wiper and toe surfaces as will be 
explained in succeeding paragraphs. Fig. 139 shows a straight sur- 
face toe moving up and down in a straight line while in Fig. 146 a 
similarly moving toe has a curved working surface. In both there is 
pure rolling action. Likewise, in Figs. 142 and 145 the working 
surface of the follower arm is straight in one case and curved in the 
other, yet in both cases there is pure rolling action. In all cases of 
pure rolling action on flat or curved surfaces it is impossible to assign 
various intermediate velocities to the follower as part of the data of 
the problem. 

329. The principle of pure rolling action between cam 
surfaces. It is a fundamental principle of pure rolling action 
between two rotating surfaces that the point of contact between them 
must always be on the line of centers. This is illustrated in Fig. 141 
where the point of contact, C, is on the line of centers A B, and where 
the contact point between the curves C D and C E will always be on 
the line of centers. This principle also applies in Fig. 139, where the 
follower toe B D is moving up and down in a straight line and where 
it must be considered that the toe is turning about a point on the line 



MISCELLANEOUS CAM ACTIONS AND CONSTRUCTIONS 109 

A B F a.t an infinite distance. Then A F becomes the line of centers 
and the point of contact between B C and B D will always be on the 
line B F. 

330. Well-known curves that lend themselves readily 
to pure rolling action in cam work are the logarithmic spiral and 
the ellipse. Examples of these will bo given in following paragraphs, 
where the solutions are entirely graphical and comparatively simple. 
The parabola and the hyperbola may also be readily used for rolling 
cam surfaces. Any line or curve that may readily be expressed by a 
mathematical equation may also be taken as one surface and the 
equation for the other curve that will work with it in pure rolling 
action may be derived. An example of this is given in paragraph 3 1<>. 
The use of the logarithmic curve for pure rolling action in the toe- 
and-wiper type of construction where the follower toe has a straight- 
edge working surface and moves in a straight line is given in the 
paragraphs immediately following. 

331. Problem 34. Pure rolling with flat surface follower. 
Required an oscillating logarithmic cam arm that will give a straight- 
line reciprocating motion to a flat-surface follower arm, with pure 
rolling action: 

(a) The follower to move up 434 units, while the cam turns 30°. 

(b) The pressure angle to be 20°. 

332. This problem is illustrated in Fig. 139 where the flat-surface 
toe B D is moved from the solid-line position to the dash-line position 
while the cam ABC swings through the angle C A F. The method 
of constructing the problem is as follows : 

Draw the horizontal line A F, Fig. 139, and from any point B 
draw a line B D making an angle with B F equal to the assigned 
pressure angle. Continue B D until the vertical distance between it 
and B F is equal to the assigned lift of the follower, 434 units in this 
problem as measured at D F. Mark the point F. Assume the dis- 
tance A B sufficient to allow for the cam shaft and cam hub. A B 
is taken as 4 units in this problem, and A F is found upon measuring, 
to be 1G units. Substitute these values in the following general 
equation : 

_ 180° X tana R = 180° X .364 _ 

~ tt X .434 l0 ^ r 3.14 X .434 X A) ° 2 " 28 ' 8 
in which r = 4, R = 16, a = 20°, and in which 6 gives the angle 
whose limiting radial line A C is equal in length to A F : 



170 



CAMS 



333. The angle of 28.8° is then laid off at F A C as shown in Fig. 
139 by means of a protractor. If a protractor is not at hand this 
angle may be readily constructed with the aid of a trigonometrical 
table from which the tangent of 28.8° is found to be .55. Lay off 
A E equal to one unit on any independent scale and draw a perpen- 
dicular line E H at E. On this line lay off .55 of this unit thus obtain- 
ing the point H. The angle E AH will then be 28.8°. Draw A H 
and continue it to A C making A C = A F = 16 units on the scale 




Fig. 139. — Problem 34, Oscillating Cam with Pure Rolling Action on Flat Sur- 
face Follower 



of the cam drawing. The logarithmic curve through B and C will 
be the one which will work in pure rolling action with the straight 
line B D. 

334. To obtain other points on the curve B C as at J, assume 
intermediate values for R in the above formula, r remaining the same 
as before. Taking R at 14 units and again solving the equation, 
is found to be 26° and this angle is laid off at F A J. A J is made 
14 units in length. In like manner other points shown by dots 
between J and B may be found by taking R equal to 12, 10, 8, and 6 



MISCELLANEOUS CAM ACTIONS AND CONSTRUCTIONS 1<I 

in successive computations and laying off the resulting angles which 
are found to be 22.9°, 19.1°, 14.4°, and 8.45° respectively. 

335. The pressure angle between the two earn surfaces will be a 
constant and equal to a. The smaller the pressure angle, the longer 
will be the toe of the follower for a given lift. As a corollary to the 
conditions of pure rolling action it follows that the developed length 
of the logarithmic arc B C must be equal to the length of the straight 
line B D. The stem E G of the follower toe may, in general, be taken 
with its center line midway between B and F. 

336. Exercise problem 34a. Pure rolling with flat sur- 
face follower. Required an oscillating cam arm that will give a 
straight-line reciprocating motion to a flat-surface follower arm, 
with a pure rolling action : 

(a) The follower to move up 1 unit while the cam arm turns 30°. 

(b) The pressure angle to be 15°. 

337. The use of the logarithmic curve for pure rolling 
action between two rolling cam arms, where both arms oscillate, is 
shown in the paragraphs immediately following. Before taking up a 
definite problem it is necessary to consider, in order to obtain a 
satisfactory understanding, some of the properties peculiar to the 
logarithmic curve. These properties are: 

First. That in a series of equally spaced radial lines drawn from 
the pole to the logarithmic curve, the length of any one line is a mean 
proportional of the lines on either side. To illustrate, the curve G H, 
Fig. 140, is a logarithmic curve, the radial lines A G, A K, A L, and 
A H are spaced by equal angles and A K : A L : : AL : A H , 
or, A L = V A K X AH. The spacing angle may be of any size. 

Second. That the difference in length between any two radial 
lines drawn from the pole to the curve will be the same no matter 
where those radial lines are taken, providing they intercept equal 
lengths of arc. To illustrate the difference C\ A — E A, Fig. 140, 
is equal to D A — C A for the reason that the arcs C D and E C\, 
were made equal in developed length. 

Third. That a tangent and a radial line at any point on a loga- 
rithmic curve form the same size of angle, no matter where the point 
is taken. To illustrate, the angle between the tangent Q C and 
the radial line A C, Fig. 140, equals the angle between Qi H and A H. 

338. Problem 35. Pure rolling with logarithmic curved 
cam arms. Construct a pair of pure rolling oscillating cam arms 



172 



CAMS 



with logarithmic curved surfaces, the driver swinging through 21° 
and the follower arm through one-half of that angle. 

339. The first step in the solution of Problem 35 consists in draw- 
ing a logarithmic curve of any desired curvature by assuming any 
convenient angle such as 60° as shown at K A H in Fig. 140 and any 
two lengths of lines as shown at A K and AH. K and H will then 




Fig. 140. 



-Problem 35, Basic Logarithmic Curve for Oscillating Cam Arms Hav- 
ing Pure Rolling Action 



be points on the logarithmic curve. To find an intermediate point, 
bisect the angle K A H as at A L and make A L a mean proportional 
between A K and A H in accordance with the first general principle 
of paragraph 337. To find a point to the left of K, make the angle 
K AG equal to angle L A K; then A K becomes the mean propor- 
tional, tmdAG : AK : : A K : A L, or A G = A -~. 

A Li 

340. Having constructed the general logarithmic curve as above, 
lay off an angle of 21° with the vertex at A, Fig. 140 and with the 



MISCELLANEOUS CAM ACTIONS AND CONSTRUCTIONS 173 

sides at A C and A D, or, the sides may be in any other position, 
according to the length desired for the cam arm. Make a tracing of 
of the angle CAD and of the arc C D and reproduce it at C A J) in 
Fig. Ml. Also draw the body outlines of the cam arm, and the driver 
is then completed. To find the follower, step off the arc C D, Fig. 
140, into four or six steps with the dividers and restep the distance C D 
off on another part of the logarithmic curve where the newly placed 
arc, equal to C D, will be subtended by an angle of 10^2° as specified 
in the data. The new position for the length of the arc must be found 
by trial, and in this problem it is at E C\ in Fig. 140 where the arc 
E C\ equals C D in length and the angle E A C\ equals one-half of 
CAD. The angle E A C\ and the arc E C\ are now traced on tracing 
cloth and redrawn at C B E in Fig. 141. Upon drawing the outlines 
for the arm the follower is completed. 

-a 




Fig. 141. — Swinging Cam Arms with Logarithmic Surfaces in Pure Rolling Action 

341. The angular motion of each cam depends on the positions 
on the logarithmic curve at which the equal arcs are taken. Had it 
been desired to sw r ing the shaft B through a larger angle, the loga- 
rithmic arc E Ci, Fig. 140, would have been taken lower down. When 
E C\ coincides in position with C D, the arm C B, Fig. 141, will 
swing through the same angle as the arm C A and both arms will be 
of the same length and identical in every way. 

342. Tangency of logarithmic cam surfaces. The fact that 
the two rolling cam curves C D and C E, Fig. 141, are tangent at C 
follows from the third principle laid down in paragraph 337, which 
points out that the tangents at C and C\ } Fig. 140, make the same 
angles with CA and C\A respectively. Since C and (\ come 
together on the same straight line A B in Fig. 141, the angle B C Q\ 
in that figure will equal the angle ACQ. 

343. Regulation of pressure angle where logarithmic 
rolling cams are used. Logarithmic curves of varying sizes, or 
expansion, may be used for rolling cam surfaces, but in general, the 



174 CAMS 

best results will be obtained by using curves having a large expansion. 
The expansion may be measured specifically by noting the rate of 
increase in the length of the successive radial lines which are drawn 
at equal angles with each other. The greater the expansion of the 
logarithmic curve, the smaller will be the pressure angle, or radial 
pressure on the bearings of the cam. This is shown in Fig. 141, 
where P C R is the pressure angle and C S is the radial pressure on 
the bearings. If curve C D had a greater expansion, its normal 
C R would fall nearer C P and the pressure angle would be smaller. 

344. Exercise problem 35a. Pure rolling logarithmic cam 
arms. Construct a pair of rolling oscillating cam arms with curved 
surfaces, the driver swinging through an angle of 30°, and the fol- 
lower through 20°. 

345. Derived curve for rolling cam arms. Any line or curve 
that is expressed by a mathematical equation may be taken as the 
form of an oscillating cam arm, and the equation for another curve 
that will work with it in pure rolling action may be derived. In 
the paragraphs immediately following, a cam arm with a straight 
surface is assumed and the curve that will work with it is derived. 
The derivation of the curve involves the use of calculus, but the 
results are comparatively easy to apply practically. 

346. Problem 36. The use of a derived curve for rolling 
cams. Given a straight-edge oscillating follower arm. Required a 
curved oscillating arm that will work with it with pure rolling action. 

347. In solving the above problem the following notation, illus- 
trated in Fig. 142, will be used : a = angle between the line of centers 
of the oscillating arms and the line of the straight follower surface 
for the phase in which this line, when extended, passes through the 
axis of the driver shaft. The angle a is a constant. 

b = angle turned through by the straight follower arm, at any 

phase, measured from the horizontal position. 
c = construction angle for the curved follower arm. 
L = distance between centers. 
R = working radius of follower arm at any phase. 
S = working radius of driver arm at phase corresponding to R. 

Then, sina = j-g = -^; (1) 

7? = BC Ri . /n\ 

a sin B DC sin V KJ 



MISCELLANEOUS CAM ACTIONS AND CONSTRUCTIONS 175 
S = L- R; (3) 

r _ 180° ta n a . cos 3^2 (& + a) 

" 7T X 0.4343 g sin J^ (6 - a) • ; • W 

Assuming L = 24, and 7?i = 4, we have from equation (1), 

4 

sin a = — = 0.16G8, and from a table of sines, a = 9%°. 

/IT 




Fig. 142. — Problem 36, Swinging Cam Arms with Derived Surfaces in Pure Rolling 

Action 



Then, from equations (2) and (3), for 
4 



b = 12° I? = 

l ' U .208 



19.22 and S = 24 - 19.22 = 4.78 



15°, 7^ = ^^^ = 15.45 " S = 24 - 15.45 = 8.55 



6 = 20°, R 



.2588 

4 
.342 



= 11.70 " S = 24 - 11.70 = 12.30 



Similarly, for b = 30°, R = 8.00; for 6 = 50°, R = 5.23; for 
b = 70°, R = 4.2G; and for b = 90°, R =4»ft. 
From equation (4), for 

180° X .1718 . oobH (12° + «%*) . 
' 3.1416 X .4343 B sin M (12° - 9%) ' 

= 22.G7 log ~ 9 - = 22.67 X 1.70 = 38.6°; 



176 



CAMS 



b = 15°, c = 22.67 log ^Z = 22.67 X 1.32 = 30°; 
' .045/ 

b = 20°, c = 22.67 log '^^ = 22.67 X 1.033 = 22.8°. 

Similarly for b = 30°, c = 16.5°; for b = 50°, c = 9.1°; for b = 70°. 
c = 4.2°; and for b = 90°, c = 0°. 

348. Plotting the above values of R in Fig. 142, B H = 19.22 and 
AH = 4.78; B J = 15.45; andBD= 11.70, etc. A test of the 
accuracy of the work thus far may now be made by drawing a line 
from H tangent to the circle having Ri for a radius and noting if it 
makes an angle of 12° with the line of centers. Likewise a line from 




Fig. 142. — (Duplicate) Problem 36, Swinging Cam Akms with Derived Surfaces 
in Pure Rolling Action 



D tangent to this same circle should make an angle of 20° with D B, 
etc. 

Again, plot the values of c, starting with any phase in which it 
is desired to show the cams. In this case the phase illustrated is 
for the straight cam at an angle of 20°. Lay off the angle D A P = 
22.8°. Then, starting with A P as a datum line lay off the values 
of c as found above, making the arc P Q = 38.6°, arc P T = 30°, etc. 

Finally on each of the lines A Q, A T, etc., lay off the correspond- 
ing values of S. These values have already been found to be 4.78 
and 8.55 respectively, etc. Thus the points F, V, D . . . P on the 
follower cam curve are obtained. 



MISCELLANEOUS CAM ACTIONS AND CONSTRUCTIONS 177 

349. Rolling cams useful for starting shafts gradually. 

The curve FDP is tangenl to the circular are having .1 P for a 

radius. This suggests an interesting and perhaps useful mechanical 

addition in that gear teeth might be cut on G P W as a pitch line, and 

also on Z C N' as a pitch line, thus permitting an oscillating shaft A to 

give a certain number of complete revolutions in opposite directions 

to the shaft B. In starting each cycle, the shaft B would accelerate 

gradually, and it would come to rest gently at the end of its cycle. 

The rate at which the motion of the shaft B would accelerate at start- 

H A N A 
ing is indicated by the ratios ^r^B to \ T ' . ( living the actual values 

11 B i V B 

which these ratios have in this problem, it is found that the accelera- 

4 78 20 
tion of B increases from ., ' ort to -r- or from about }A of the angular 

19.22 4 •* e 

velocity to five times that of the shaft A . 

350. Regulation of pressure angle with derived rolling 
curve. Returning to Problem 36 and considering it only as a cam 
mechanism it will be noted that the angles taken for b in the compu- 
tations become the pressure angles and show a measure of the radial 
thrust that goes into the bearings without producing any useful 
rotative effort. For example, in Fig. 142, the cams are in contact 
at D and the normal pressure is represented by D U. The component 
pressure D X goes to the bearings and D Y is useful in turning the 
shaft B. The pressure angle U D Y is 20°. When G reaches M, it 
will be in contact with Z and the pressure angle will be 50°. 

351. Exercise problem 36a. The use of a derived curve for 
rolling cams. Given a straight-edge oscillating follower arm. 
Required a curved oscillating arm that will work with it in pure 
rolling action. Let distance between centers of oscillating arms 
be 30 inches and the maximum theoretical length of the curved toe 
20 inches. 

352. Elliptical arcs for pure rolling cams. Pure rolling 
oscillating cam arms having arcs of ellipses for their working sur- 
faces, may also be used. In constructing these cams, use is made of 
the characteristic of the ellipse that the sum of the two lines drawn 
from any point on the perimeter to the foci will be constant and 
will be equal to the length of the major axis. Briefly then, it is only 
necessary to take two identical ellipses and center them on one pair 
of their foci at a distance apart equal to their major axis. Such a 
pair of ellipses, illustrated at .4 C and B D in Fig. 143, will then 



178 CAMS 

turn through 360° respectively and will be in pure rolling contact all 
the time. Oscillating rolling cam arms may be obtained from the 
ellipses by simply taking equal and symmetrically placed arcs from 
each as shown at E F and G H, Fig. 143. The following problem 
will illustrate the method of construction. 

353. Problem 37. Pure rolling elliptical cam arcs, angles 
of action equal. Construct a pair of oscillating rolling cam arms 
whose working surfaces are arcs of ellipses. Take the distance 
between centers as 24 units, make the angle of action of the driver 
and follower shafts the same, and find the pressure angle at any 
point. 

354. In constructing Problem 37, lay down the assigned dis- 
tances between centers, 24 units, as at A and B in Fig. 143. These 
points will lie at the fixed focus of each ellipse. Take any point, 
such as K, on the line of centers between A and B. The nearer 
K is taken to one of the foci the smaller will be the pressure angles 
between the rolling cam surfaces according to this construction, 
other data being the same. With K as a center and K B as a radius 
draw an arc B L of any desired length thus obtaining the angle B KL 
which may be any value. The smaller it is taken the flatter will be 
the resulting working surface G H of the cam and the smaller will be 
the pressure angles. Had K been taken midway between A and B, 
and had the angle B KL been made 90°, a limiting case would have 
resulted in which the ellipses from which the cam arms are taken 
would have had a minimum eccentricity and the cam arms would 
have had the largest angle of action, but the pressure angles would 
have been larger. With K as a center draw the arc A I making 
angle A K I equal to B KL. Then L and J are the free foci of the 
basic ellipses. 

355. To find the major and minor axes of the ellipses 
take L and A, Fig. 143, as centers and one-half of A B as a radius 
and draw short arcs intersecting at M and at N as indicated at M. 
Also use B and / as centers in the same way, thus obtaining and P. 
M and JV will then be the extremities of the minor axis of one ellipse, 
and and P the extremities of the other. From /, which is midway 
between A and L, lay off distances J Q and J C equal also to one-half 
of A B. Q and C are then the extremities of the major axis of one 
ellipse, and similarly D and R are the extremities of the other ellipse. 

356. To find points of the ellipse take a piece of paper, or 
a thin card, having a perfectly straight edge as indicated by the dash 



MISCELLANEOUS CAM ACTIONS AND CONSTRUCTIONS 179 



and double-dot lines in Fig. 143. Mark the points T and U on the 
edge of the paper a distance apart equal to the semi-minor axis S, 
and also mark the point V so that its distance from T is equal to the 




..Fig. 143. — Problem 37, Basic Ellipses for Pure Rolling Cam Arms, Angles of 

Action Equal 

semi-major axis D S. Then move the paper so as to keep the 
point U always on the major axis, and V always on the minor axis, 
and the point T will move in the path of the desired ellipse. 

357. TO OBTAIN AN EQUAL ANGLE OF ACTION FOR BOTH ELLIPTICAL 

cams, as called for in the statement of the problem, equal lengths 



180 *-' CAMS 

of arcs symmetrical about the extremity of the minor axis are taken 
from each ellipse. Thus N E equals N F, Fig. 143, and G equals 
H equals N F. The angle of action for each cam is then equal to 
E A F. This angle may be made larger or smaller by increasing or 
decreasing the arcs N F and N E. These arcs, however, should 
not approach too closely to the extremities of the major axis, for the 
pressure angle then increases rapidly, as, for example, when the con- 
tact point moves from F toward Q. 

358. Pressure angle in rolling elliptical cam arms. The 
pressure angle is the angle between the perpendicular to the radial 
line at the point of contact and the normal to the curve at that point. 
It varies at different phases and is a minimum when the extremities 
of the minor axes are in contact, that is when N and 0, Fig. 143, 
come in contact at X. Consequently the angle S Z is the pressure 
angle when is in action. The line Z is perpendicular to the radial 
line B 0, and the line S is normal to the curve at 0. At K the 
angle W K Y is the pressure angle. The normal to the ellipse at 
any point, such as K Y, may be readily found by making use of the 
property of the ellipse that the normal to the curve at any point is 
the bisector of the angle formed by the focal lines from that point. 
For example K B and K I are focal lines from K, and K Y bisects 
the angle B K I. 

359. Exercise problem 37a. Pure rolling elliptical cam 
arms, angles of action equal. Construct a pair of oscillating 
rolling cam arms whose working surfaces are arcs of ellipses. Take 
the distance between centers as 20 units, make the angle of action 
of each shaft the same, and find the pressure angle at the extremity 
of the angle of action. 

360. Problem 38. Elliptical rolling cam arcs, angles of 
action unequal. Construct a pair of oscillating rolling cam arms 
whose working surfaces are composed of an arc of an ellipse. Take 
the distance between centers as 24 units, make the angle of action of 
the driver 2.9 times that of the follower and find the maximum 
pressure angle. 

361. In the solution of this particular problem any ellipse may be 
used whose major axis is 24 units long. The shorter the minor axis 
is taken the less will be the pressure angle, and the smaller also will 
be the actual practical angles through which the cam arms will 
swing. In laying down the problem take Q C, Fig. 144, equal 24 
units, as the major axis of the ellipse. Bisect Q C at X, and assume 



MISCELLANEOUS CAM ACTIONS AND CONSTRUCTIONS 181 

X M and X N as the semi-minor axes. With M and N as centers 
and a radius equal to Q X draw short arcs intersecting the major axis 
at A and at L. These points will be the foci of the ellipse. ( 'oust ruct 
the ellipse as directed in paragraphs 355 and 356. Select an arc of such 
length and position on the ellipse that it will subtend focal angles, i.e. 




Fig. 144. — Angles of Action for Elliptical Cam Arms 



angles whose vertices are at the foci, which are to each other as 1 is to 
2.9. Such an arc is shown at F E and it subtends an angle of 24° 
from the vertex at L and an angle of 70° from the vertex at A. The 
value of 2.9 given in the data is now provided for because 70 divided 
by 24 equals 2.9. The arc F E is used for the form of the working 
surface of the two cam arms, as directed in the following paragraph. 
362. To construct the cam arms for Problem 38, lay down the 
shaft centers by marking the points A and B, Fig. 145, 24 units apart. 
On a piece of tracing paper draw the arc F E of Fig. 144 and mark 
the point A. Lay the tracing paper down in Fig. 145 with A of the 




Fig. 145. — Problem 38, Pure Rolling Elliptical Cam Arms, Angles of Action 

Unequal 



tracing at A of the figure, and with E of the tracing on the center 
line A B. Redraw the traced curve in Fig. 145, giving E F. Again, 
on the tracing paper draw the curve F E of Fig. 144 and mark the 
point L. Lay the tracing paper down in Fig. 145 with L of the 



182 CAMS 

tracing at B, and E of tracing on the center line A B. Redraw the 
traced curve in Fig. 145, giving E H. E F and E H should be tan- 
gent at E if the work is correctly done. The forms of the arms and 
the hubs are now drawn, and the angles of 70° and 24° are indicated 
as shown in Fig. 145, thus giving a ratio of turning angles of 2.9 to 1 
as required. The maximum pressure angle will be at the point on 
the working curve that is nearest to the extremity of the major axis 
of the original ellipse, and it will be equal to 50° as shown at W F Y, 
Fig. 144. 

363. Exercise problem 38a. Elliptical rolling cam arcs, 
angles of action unequal. Construct a pair of oscillating rolling 
cam arms whose working surfaces are composed of an arc of an ellipse. 
Take the distance between centers as 18 units, make the angle of 
action of one shaft 2.2 times that of the other, and find the maximum 
pressure angle. 

364. Pure rolling parabolic cam surfaces for a recipro- 
cating motion. The parabola lends itself to pure rolling action in 
cam work, but it can be used only when either the driver or the fol- 
lower has rectilinear motion, and then the rectilinear motion must 
be in a direction perpendicular to the line of axes of the two para- 
bolas when they are in contact at their vertices. 

365. Problem 39. Rolling parabolas. Required a parabolic 
oscillating cam to give rectilinear motion to the follower with pure 
rolling action. Assume the length of path of contact, and find, (a) 
angle of action of driver, (b) range of motion of follower, and, (c) the 
maximum and minimum pressure angles. 

366. Construct a parabola by making use of the property that 
a point on the curve is equidistant from the focus and the directrix. 
To do this, assume a point A, Fig. 146, as the focus of the parabola 
on the line XX as an axis. Assume the directrix Y Y at right 
angles to the axis and at any desired distance A B from the focus. 
The larger A B is taken the larger will be the oscillating cam for a 
given motion, and the smaller will be the pressure angles. The 
vertex of the parabola will be at J midway between A and B. A 
point on the curve may be found by assuming any radius, such as 
A D, and drawing a short arc as shown at D using A as a center; 
then laying off this radial distance on the axis starting from the direc- 
trix as &t B Di, and drawing a perpendicular line D\ D until it meets 
the arc at D. The point D, thus obtained, will then be a point on 
the parabola and other points may be found in the same way and the 



MISCELLANEOUS CAM ACTIONS AND CONSTRUCTIONS 183 

curve S J V drawn. Draw a radial line A K perpendicular to the 
desired direction of motion of the follower, which direction, in this 
problem, is C W. On this radial line, assume any distance R K 
as the path of action ; then, arcs of circles through R and K having A 
for a center will limit the part G E of the parabolic curve which will 
form the driver cam surface. The entire oscillating cam G E A may 
now be drawn. 




4 _____ 










L^- 






\ ^ 






. 






/^Vl 


X 


D, 




A n" 




fliC = Path c 


f Action 







Fig. 14G. — Problem 39, Parabolic Cam Surfaces for Pure Rolling Reciprocating 

Motion 



367. The surface ilfiV on the follower arm, Fig. 146, will be 
identical with G E, already constructed, and may be readily found by 
drawing the lines SJV and X X on a piece of tracing cloth or paper, 
turning the paper on the reverse side, and then adjusting it, always 
keeping X X of the tracing parallel to A K, until the two curves come 
tangent as shown at C. The axis X X of the tracing will then be in 
the position Z Z. The follower cam surface and rod, M N U may 
now be drawn. The angle of action, range of action, and pressure 
angles may now be found as indicated in Fig. 146. 



184 CAMS 

The two parabolic surfaces G E and M N, Fig. 146, will be in 
pure rolling action on the path K R, the driving cam turning about A, 
and the follower cam moving in a direction perpendicular to K A . 

368. Pure rolling hyperbolic cam arms where centers are 
close together. Two equal hyperbolas will give pure rolling 
action to two oscillating cam arms, the essential features of con- 
struction being that the hyperbolas should turn about one pair of 
foci as fixed centers and that the distance between these centers 
should equal the distance between the vertices of the hyperbolas. 
In Fig. 147 A and B are the foci of one hyperbola and C and D U 
its two branches, while H and S are the foci of the other hyperbola 
and V W one of its branches. 

369. Problem 40. Rolling hyperbolas. As a problem illus- 
trating the application of hyperbolas to rolling cam work, let it be 
required to construct two cam arms and shafts and determine the 
angle of action of each and the maximum and minimum pressure 
angles. 

370. Construct the hyperbolas by making use of the property 
that the distances from any point on the curve to two fixed points, 
called the foci, have a common difference. Therefore, assuming A 
and B as foci, Fig. 147, and C as a vertex, the common difference to 
be used throughout will be C B minus C A, equals C D. By assuming 
different distances between A and B, and A and C, different angles of 
action and different pressu e angles will be obtained. 

371. A point on the hyperbola, such as E, Fig. 147, is found by 
taking any radius such as B E and striking an arc with B as a center. 
Then with A as a center draw another short arc with a radius equal 
to B E minus C D. Where the second arc crosses the first will be 
a point on the curve as at E. Other points are found in the same 
way. 

372. The center of one shaft will be located at the focus B if it is 
desired, for example, to show the cam surfaces in contact on the 
branch C 0. Assuming E as the point of tangency of the two cam 
surfaces, the center of the other shaft, and consequently one focus 
of the other hyperbola, must be on the line E B, for the reason that 
in pure rolling action the point of contact must always be on the line 
of centers. 

373. The second hyperbola must be placed with respect to the 
first so that the distances between the fixed foci and the free foci are 
equal to each other and to the distance between the vertices of the 



MISCELLANEOUS CAM ACTIONS AND CONSTRUCTIONS 185 

hyperbola. Therefore, take the distance C D and lay it off at 
B H, also use it as a radius with A as a center to draw the short arc 
through S. With H as a center and A B as a radius draw another 
arc intersecting the first at S, thus determining the second focus, S, 
of the second hyperbola and its axis if it is desired. The tangent 
branch V W may now be independently constructed as explained 
for C 0, or it may be traced from C of which it is a duplicate. 

Q 




Fig. 147. — Problem 40, Hyperbolic Cam Surfaces for Swinging Arms on Close 
Centers with Pure Rolling Action 



374. The path of action, assuming K G to be the driving cam 
surface, the angles of action for both cam shafts, and the maximum 
and minimum pressure angles may be obtained from a study of the 
illustration in Fig. 147. 

Rolling hyperbolic cams differ from all others in that the path of 
action may lie entirely on one side of the centers of rotation, instead 
of between the centers as is the case with the logarithmic and ellip- 



186 CAMS 

tical arms. The practical value of this is that it will permit driving 
action between two shafts that are very close together, as indicated 
by the shafts H and B in Fig. 147. 

375. Detail drawing of cylindrical cams. A simple practical 
method for constructing the surface guide line for the center of the 
cutting tool in cylindrical cams was explained in paragraph 127. A 
more elaborate method of construction giving a more precise mechan- 
ical action and a more complete drawing of the cam is now given. 

376. TO FIND THE TRUE MAXIMUM PRESSURE ANGLE OF A CYLIN- 
DRICAL cam, the pitch cylinder and not the surface cylinder should 
be drawn first. The pitch cylinder is shown at B' W Hz . . . Qs, 
Fig. 148 and is drawn with a radius of 5.2, the data, excepting for 
pressure angle, being the same as for Problem 15. Briefly the data 
are: (a) Follower to move in a straight line 4 units to the right on 
the crank curve while the cam turns 120°; (b) To move to left 4 
units on crank curve while cam turns 120°; (c) To dwell while cam 
turns 120°; (d) the true maximum pressure angle to be 30°. 

The pitch curve Q2 P2 J, etc., is then obtained and the normal 
J G\ is drawn as in Problem 15. This normal will make an angle of 
30° with J D if the work is correctly done. This is the assigned 
maximum pressure angle and is at the bottom of the pin; the pressure 
angle D J G at the surface of the cylinder will be less than 30°. 
In this problem the data and layout were such that the point J of 
maximum pressure angle could be readily made to fall on the front 
element of the cylinder and the angle D J G\ thus shown in its true 
size. Where the data and layout are such as not to conveniently 
bring the pitch point on the front element of the cylinder, the pitch 
curve will have to be revolved if it is desired accurately to show the 
pressure angle in true size on the drawing. 

377. Drawing of groove outlines for cylindrical cam. If 
it is desired to draw the groove outlines of a cylindrical cam one of two 
methods may be used, (1) the approximate method, or, (2) a more 
exact method. The approximate method which is simpler and 
quicker and which gives good-appearing results where the slope of 
the groove does not exceed 30° is applied by laying off the points 1 
and 2, Fig. 148, at equal distances on each side of H2 on the surface 
pitch curve, these points representing the extremities of a diameter of 
the follower pin. Similarly the points 8-4, 5-6, etc., are obtained. 
A curve drawn through the points S, 1, 3, 5, etc., will represent one 
of the surface edges of the groove. The bottom lines of the groove 



MISCELLANEOUS CAM ACTIONS AND CONSTRUCTIONS 



1S7 



are found, for example, by projecting J5 to J a and then laying out 

the diametral line 9-10. A curve through 9 iwA other similarly 




o 
< 

a 
a 
P 
o 

t/3 

a 

« 

B 
P CD 

« a 

^ s 



5 O 
5 o 

j a 



found points will represent one of the bottom edges, and the curve 
through 10, etc., will represent the other. 

378. A MORE EXACT METHOD OF DRAWING THE OUTLINES OF THE 

groove consists in drawing the projection of a section of the pin 



188 CAMS 

which is tangent to the cylinder. The section will appear in general 
as an ellipse in the side view and the curves representing the groove 
edges will be drawn tangent to these ellipses instead of through the 
extremities of the major axes as described in the preceding paragraph. 
The detail work for this is shown in Fig. 148 where the straight line 
7' -8' is the end projection of a pin section which is tangent to the 
cylinder. The points 7 and 8 are projected from 7' and 8' and are 
the extremities of the minor axis of the ellipse; the horizontal line, 
5-6 passing through J 2 is equal to the diameter of the pin and is 
equal to the major axis. J 2 is projected from J'. The ellipse 
5, 7, 6, 8, is now constructed, as shown. Similar ellipses should be 
constructed at other points as at 1 2, #2, etc. At A the ellipse 
becomes a circle and at E it flattens to a straight line. The curve 
S E2 drawn tangent to the ellipses instead of through the extremities 
of the major axes will be one of the surface edges of the groove. 
Even with this refined method of construction there remains an 
approximation, for it will be evident that the circle at the top of the 
pin lies in a plane which is tangent to the cylinder and that the pro- 
jection of this circle gives an ellipse that does not lie on the surface 
#f the cylinder. Therefore, the curve drawn tangent to the ellipse 
would not lie on the cylinder. The error, however, in following the 
above directions is too small to show in a drawing of practical pro- 
portions. If desired, this slight error in construction may be cor- 
rected by rounding off the end of the pin to conform with the curve 
of the cylinder and projecting the curve of the rounded end of the 
pin to the side view to give the elliptical-like curve to which the 
groove curve is tangent. This is illustrated in a case of exaggerated 
proportions in Fig. 149 where M is a true ellipse and is a projection of 
the end of the pin when it is flat. N is a projection of the perimeter 
of the pin when its end is rounded off to conform with the curve of 
the cylinder. 

379. Forms of follower pins for cylindrical cams. Cylin- 
ders, cones and hyperboloids may be used for the form of follower 
pins to work in the groove of cylindrical cams. A cylindrical pin, 
drawn to a large scale, is shown at G J in Fig. 149 lying in a groove 
which is cut in the cam cylinder C Z. The cylindrical pin is advanced 
longitudinally the distance E F, Fig. 149, while the cam turns through 
the angle A\ 0\ A 5, Fig. 150. The top edges of the groove are rep- 
resented by the helical curves G G\ and H H4, Fig. 149, and the bot- 
tom lines of the working side surfaces of the groove are represented 



MCSCELLANEOUS CAM ACTIONS AND CONSTRUCTIONS 189 

by 1 1\ and J J \. The center of the follower pin is shown in its 
central position at At.. The straight line A% G2 is a normal to the top 
pitch line A A2 A± of the groove, and it is the line of pressure between 
the side of the groove and the follower pin at the surface of the 
cylinder. The angle K A2O is the pressure angle at the top of the 
pin and it is made 30° in this example as shown in Fig. 149. The 
straight line A2 1 2 is a normal to the helix B A 2 B± which is the locus 
of the center point of the bottom of the follower pin. The line A2 h, 
then, is the line of pressure between the side of the groove and the 
pin at the bottom of the pin, and L A 2 is the pressure angle at the 
bottom of the pin. The pressure angle, therefore, varies from the 
top to the bottom of the groove, being smallest at the top. From 
this it follows that the pitch surface of a cylindrical cam should be 
at the shortest radius reached by the follower pin rather than 
at the outer surface where it is usually taken, provided it is 
desired not to exceed a given maximum pressure angle on the 
follower pin. 

380. When the pin is moving, the line of contact between 
the side of the groove and the side of a cylindrical pin is a curved line, 
and one phase is shown in end projection at Go 1 2 in Fig. 149 and in 
side projection at Gi h in Fig. 150. When the pin is not moving 
it has straight-line contact with the side of the groove as shown 
at G I in Fig. 149. If the follower pin is fixed in the frame that 
carries it, it will receive wear on the forward stroke entirely within 
the area G Gq Iq I G, Fig. 149. Gq is the same horizontal distance 
from the vertical centerline through A as G2 is from the vertical 
centerline through A2. 

381. A rotating cylindrical pin cannot have pure rolling 
action against the side of the groove in a cylindrical cam, for 
such action requires at least that two rolling curves must measure 
off their lengths equally on each other. This means that if the cir- 
cumference of the follower pin at the top goes a certain number of 
times in the curve G G2 G±, the circumference at the bottom must go 
the same number of times into the curve / 12 I±. This, of course, 
cannot happen with a cylindrical pin for the circumferences of the 
pin are the same top and bottom while the helical curves at the top 
and bottom of the groove against which they operate are totally 
different in length. From the above it follows that there will be 
considerable sliding between a cylindrical pin and cylindrical cam 
and that this will be greater, the greater the length of the pin. 



190 CAMS 

382. A conical follower pin for a cylindrical cam is shown at 
M\ Ri in Fig. 150 and in end view in Fig. 151. In the latter view the 
line G Gi is tangent to the helix which marks the center of the top of 
the groove and A2 G2 is normal to it, giving the point G2 at which the 
conical pin is tangent to the side of the groove at the outer circum- 
ference. The conical pin is here taken the same size at the top as the 
cylindrical pin in Fig. 149 and consequently the line A 2 G2 in Fig. 151 
will be parallel and equal to the line A2G2 in Fig. 149. Likewise 
I2, Fig. 151, is the point of tangency at the inner end of the pin. 
These points of tangency, and intermediate ones, will determine the 
line of contact G2 I2 between the conical pin and the side of the 
groove for the position shown. This line is also shown in side pro- 
jection at G3 1 3 in Fig. 150. If the pin is rigidly attached to the fol- 
lower framework the wear on the pin will fall on the area represented 
by the surface Si S3 1 3 G3. If the pin is free to turn on its axis it will 
come nearest to having rolling action when the circumference at the 
bottom of the conical pin is to the circumference at the top as the 
length of the helix B B± is to the helix A A\ in Fig. 149, or, when the 
conical vertex of the roller is at Oi, Fig. 150, on the centerline of 
the cylinder. Conical pins give thrust in an axial direction and conse- 
quently there must be special provision in the follower framework 
for holding the pin in place. Conical pins have a natural advantage 
in that they may be designed to move in axially and so to take up 
wear in the pin and in the groove. 

383. An hyperboloidal follower pin is shown at T\ Vi Y\ W\ 
in Fig. 150 and in end view in Fig. 152. In the latter Figure the 
lines A2G2 and A2I2 are perpendicular to the top and bottom helices 
of the groove respectively, the same as the similarly lettered lines in 
Fig. 149. If the diameters T\ Wi and Ui Vi are taken in the same 
ratio to each other as the lengths of the top and bottom helices of the 
groove in which the pin rolls the closest approximation to rolling 
action will be obtained. There cannot be pure rolling of the hyper- 
boloidal pin, however, on the side of the groove, for, even where the 
circular sections of an hyperboloidal pin measure themselves off 
equally on the corresponding helices on the surface against which 
they roll, there is always an inherent fundamental endwise or longi- 
tudinal component of sliding in the direction of the axis of the 
pin in every hyperloidal action. The nature and amount of this 
characteristic is explained in some of the books on descriptive 
geometry. 



MISCELLANEOUS CAM ACTIONS AND CONSTRUCTIONS 191 




192 



CAMS 



384. To lay out the hyperboloidal pin the lines A2G2 and 
A2 h, Fig 152, and the circles T W and U V are drawn. The straight 
line from G2 to h is drawn and used as the generatrix of the hyper- 
boloid, the outline of which is the curved line T\ U\ in Fig. 150. 




Points on this curve are found by dividing G2 I2, Fig. 152, into say 
four equal parts, revolving the dividing points to the line T U and 
then projecting them to the equally spaced lines which are drawn 
from Si to $3 in Fig. 150. The straight line (74 I± will be the line of 



MISCELLANEOUS CAM ACTIONS AND CONSTRICTIONS 193 

contact between the pin and the groove at the maximum pressure 
angle, and the curved hyperboloidal Line S\ S3 the line of contact at 
the end of the stroke. The wear on a pin fixed in the follower frame 
would occur on the hyperboloidal surface between these two lines. 

385. Plates for cylindrical cams. Instead of actually cutting 
grooves in cylinders, flat cam plates are often formed and then 
curved to the diameter of a cylinder and screwed on. The follower 
pin then w r orks against the edge of the formed plate. Such a case, 
illustrated from the working drawings of an automatic machine, is 
given in Figs. 153 and 154. Fig. 153 is a development of the plate 
as first laid out, and Fig. 154 an end view of the plate after it is 
curved and ready to be applied to the blank cylinder by fastening 




Fig. 154. — Cam Plate Formed to Fit Blank Cylinder 

screws as indicated. The developed cam surface, as shown in this 
illustration, has a straight line and gives uniform velocity to the 
follower pin. 

386. Adjustable cylindrical cams in automatic work for 
processes involving varying sizes of product. By making a 
series of plates of varying proportions, but curved to fit a single 
pitch cylinder which remains permanently in place, and by fastening 
the proper series of plates to the cylinder for a given job, the same 
automatic machine is made to turn out products of varying sizes 
without removing cam bodies from shafts. A type of automatic 
machine in which such curved cam plates are used is the screw man- 
ufacturing machine. A diagram of a blank cylindrical cam with 
curved cam plates fastened on is shown in Fig. 11. 

387. Double-screw cylindrical cam. This is the cylindrical 
grooved cam with specially formed follower pin or head adapted to 
give long ranges of reciprocating motion to a follower bar. The 
cam may be constructed to give uniform or variable velocity to the 
follower throughout its entire range of motion; also to give periods 
of rest at the end of the stroke, if desired, ranging anywhere from zero 



194 



CAMS 



to periods measured by the time required for the cam to make about 
l 2 /3 turns. The form of the follower pin should be as indicated in 
general way in Fig. 155 being elongated at J, Ji, so as to take the 
open space at two intersecting grooves and at the same time so as 
not to bind at the reverse position at the end of the stroke as at C, 
Fig. 155. 

388. If the follower motion is uniform throughout the entire 
length of both strokes, the right and left grooves C D M and L F C 
respectively in Fig. 155 will have a true helix for their pitch line. 
The angle at the end of the screw formed by the intersecting helices 
may be eased off as shown in Fig. 157, by the change from C\ Y to 
R Y. If it is desired to have the follower pause at the end of the 
stroke, the helical groove will run into a groove whose sides are 



QEZ3 




M IF 

Fig. 155. — Double Screw Cam with Half Turn Stop 



bounded by planes perpendicular to the axis of the cylindrical cam, 
as shown at L M, Fig. 155. The length of the stop of the follower 
will be equal to the time required for the cam to make one-half a 
turn in this case, as shown in Fig. 155, and to about % turn if made as 
shown at C D E F in Fig. 156. This is about the practical limit for 
time of stop for a simple groove construction. 

389. Periods of rest corresponding to more than one rev- 
olution of the cam may be obtained by a special attachment 
shown at H N in Fig. 156. A movable triangular piece, H, turns 
through an angle of a few degrees on a pin J which is mounted on a 
small sliding block that is constantly pressed to the left by the spring 
N. If the follower head V K is imagined to move around the cam, it 
will, shortly after passing M press the tip S of the piece H causing it 
to turn slightly on the pin J and so leaving a full vertical opening at 
T which the follower head will enter after passing L on its way 
around. The follower head, having entered the large open end of 
the groove at T S will move the block H to the right against the 



MISCELLANEOUS CAM ACTIONS AND CONSTRUCTIONS 195 

spring pressure at N as it passes; and as the follower head moves 
toward and past S, the spring N will cause the tip T of the block to 
move to the left and so open full the inclined groove L M I for the 
follower head to enter the next time it comes around. 




Fig. 156.— Double Screw Cam with | and If Turn Stops 



390. A SLOW ADVANCE OF THE FOLLOWER AND A QUICK RETURN 

may be obtained with the double screw cam, as shown in Fig. 157, 
when the time for the travel of the follower bar P K Q to the right is 
represented by two turns of the cam, while the return motion to the 
left will be made in one turn of the cam as shown by the increased 
(double) pitch of the helical groove R F L. To accurately represent 
the helical grooves where the pitches are different, as in this case, it 



m? 


i: 


— » 3»- 




Qm 


s 


! ! 








m 


1 1 




.F 


m 




Fig. 157.— Double Screw Cam for Slow Advance and Quick Return of Follower 

must be kept in mind that the normal distance between helices must 
always be equal to the effective diameter of the follower pin and 
that this will give a wider groove, measured parallel to the axis, when 
the pitch is large than when it is small. This is shown at E, Fig. 
157, where the large pitch and small pitch grooves are tangent to 
the'circle on the follower head. The pitch of the helix or groove is 
the longitudinal or axial advance of the follower in one full turn oi 
the cylinder. It is shown in Fig. 155 by the distance D 0. 

391. The helical groove shown in Fig. 155 gives uniform motion 
to the follower, and the helical curves which bound the groove are 



196 



CAMS 



constructed in the same way as described in Problem 15, paragraph 
126 et seq. except that the horizontal spaces AH, HI, and / J, 
Fig. 57, are made equal for the helix construction. Any practical 
variable motion may be obtained with this type of cam by varying 
the inclination of the grooves, the smoothest action for driving the 
follower from one end of the stroke to the other regardless of inter- 
vening velocities being the parabola curve. 

392. Straight sliding plate cams. The sliding plate cam M N, 
Fig. 158, is but a simpler form of the rotating cam. The figure illus- 
trates two common uses of the sliding cam, the lifting rod, A C, on 
the left operating a poppet valve, and the bellcrank B E F on the 
right a belt shifter. The general information necessary to lay out this 



y////////////////////////^^^^ 




Fig. 158. — Sliding Plate Cam. 



cam is explained in detail in Problems 3 and 16 respectively. In the 
' present case let it be required that a poppet valve be lifted the dis- 
tance 2 units with uniform acceleration and retardation with a maxi- 
mum pressure angle of 30°; find the length and form of cam surface. 
The cam factor for the parabola, which gives uniform acceleration 
and retardation, is 3.46 and, therefore, the length of the sliding cam 
surface will be 2 X 3.46 = 6.92 as shown in Fig. 158, giving a pres- 
sure angle of 30° at G. The curve A G B is laid out in the regular 
way for the parabola, by dividing H G into, say, 16 equal parts and 
taking the 1st, 4th, 9th and 16th parts and drawing horizontal lines 
drawn through them as indicated. K G is divided in the same way. 
A L is then divided into 8 equal parts and vertical lines drawn at 
each of the points to meet the horizontal lines as at P, Q, etc. 

393. The bellcrank B E F, Fig. 158, will have approximately 
uniform angular acceleration and retardation if the cam surface, 



MISCELLANEOUS CAM ACTIONS AND CONSTRUCTIONS 197 

obtained as above is used, but if exact angular acceleration is required 
the method described in paragraph 136 et seq., for the cylindrical 
cam chart must be followed. In this case the cam chart becomes 
the sliding plate cam. 

394. Involute cam. The involute curve may be used for cam 
outlines. It gives a characteristic motion almost identical with the 
cam having a straight-line base curve (paragraph 32), but it is not so 
simple to construct. The involute cam will not give true motion to a 
roller follower unless the ends of the cam working surface are eased 
off, as they are in the straight-line combination cam, or the logarith- 
mic combination cam (paragraphs 33, 56, 59, and 199) by arcs of 
c'rcles or other curves. For the same data as were taken for com- 
parison of cams shown in Figs. 71, 75, 79, etc., the involute curve 
gives a slightly larger cam than does the straight-line base curve, 
the maximum radius being 2.78 for the former and 2.65 for the 
latter. The method of finding the maximum radius for the involute 
cam will be explained in the next problem. 

395. The involute is popularly defined as the curve that 
would be generated by a point on a string which is being unwound 
from the periphery of a circular disk, the string always being kept 
taut and always in the same plane as the disk. 

396. The involute is readily constructed, according to the 
preceding definition, by drawing a circle of any size, as illustrated at 
S P R, Fig. 159; taking any point as S as the origin of the involute; 
laying off a series of equal angles, of any desired unit, as at S M, 
M N, etc. ; drawing tangents to the circle at M , N, etc. ; and mak- 
ing the lengths M Y, N U of the tangents equal, successively, to the 
lengths of the arcs MS, N S, etc. This latter operation may be 
done graphically by setting the small dividers to a step of 34 mc h or 
less, starting at S and counting the steps toward M until M is reached 
or passed, and then counting off the same number of steps in the 
reverse direction going along the tangent line, thus obtaining the point 
Y on the involute curve. This graphical method of stepping off dis- 
tances, although generally used, is apt to give an appreciable cumu- 
lative error, and therefore should be checked by a simple computa- 
tion as follows: Length of tangent TV U, for example, equals length 

, Ar e , OSX2X 3.14 X angle NO Sin degrees T 

of arc N S, equals - * — . In 

360 

general it is advisable to first compute and draw a long tangential 

line as R W at 180° from S, and then if six equally spaced construe- 



198 



CAMS 



tion points are used as at M, N, etc., to make the tangent P V one- 
half of R W; the tangent at M Y, one-third of P V; the tangent at 
N U, two-thirds of P V, etc. 

397. Pressure angle with involute cam. Pressure angle is 
defined as the angle made by the line of action of the follower and the 
normal to the pitch curve of the cam. Therefore if the follower moves 
in the direction V, Fig. 159, and if the normal to the involute at the 
point V is V H, the pressure angle is H V K. The angle H V K 
grows smaller as the point V is moved to the right towards W, and 




Fig. 159. — Involute Curve as Used Specifically in Cam Construction 



larger as it is moved to the left towards the origin of the curve at S. 
At S the pressure angle would be 90° because the involute is tangent to 
the line of action S of the follower. The line of action of the follower 
is a radial line in the type of cam being considered in this problem. 
From the above it may be seen that there are a series of points on 
the involute where there are definite pressure angles, and these points 
will be noted here as they are necessary in solving a specific problem. 
398. At E, Fig. 160, the pressure angle is 20°. The point E is 
obtained by laying off an angle of 88° from the origin of the curve, as 
at $ E. The other points for pressure angles of 30°, 40°, etc., are 
found at A, D, etc., by using the values given in the following table. 



MISCELLANEOUS CAM ACTIONS AND CONSTRUCTIONS 199 

The method of finding these values will be given in ;i later paragraph 
for those who may be interested. 



Pressure angle. 
Initial angle. . . 



20 c 



30° 40° 
39^° 18° 



50 ( 



60° 
3° 



399. Problem 41. Required an involute cam that will move 
a radial follower one unit while the cam turns 60° with a maximum 
pressure of 30°. 

400. To solve this and any other similar involute cam problem 
it is necessary to construct first an accurate basic involute curve of 
any convenient size as directed in the preceding paragraphs; then 
to lay off the initial angle corresponding to the given pressure angle 




Fig. 160. — Showing that All Si^es of Involute have the Same Initial Angle for 

Each Pressure Angle 



in the problem. In this problem the pressure angle is given as 30°, 
for which the initial angle is 39J^° as determined from the table, 
and this latter angle is laid off at S A in Fig. 159 where the basic 
curve has been drawn. From A lay off the given working angle of 
60° as at A F. Draw the circular arc A G and measure the dis- 
tance G F. Then make a proportion in which the distance G F is to 
the assigned follower motion as the radius A is to the desired short- 
est radius of the pitch surface of the cam. In this problem GF 
measures 1.12 units, the assigned follower motion is 1 unit, and the 
radius A = 2.00 units. 
Therefore, 



1.12 : 1.00 :: 2.00 : x. 



or 



1.78, 



200 



CAMS 



equals the shortest pitch radius of the cam. With this value known 
it is a simple matter to draw the required involute cam as in Fig. 161, 
where A equals x as just computed, H A K equals the pressure 
angle of 30°, and the line H A extended is tangent to the base circle 
E Q, which is necessary in the drawing of the desired involute. With 
these items laid down, the involute pitch surface from A to F is con- 
structed in detail as described above in the paragraph devoted to the 
construction of the involute, and briefly, in review, as indicated in 
the following paragraph. 

401. Lay off the assigned working angle of 60° as at A O F, 
Fig. 161. Make F equal to A plus the assigned motion of the 




Fig. 161. — Problem 41, Involute Cam for Radial Follower Based on the Specific 

Data 



follower which is one unit in this problem. Draw F Q tangent to the 
base circle. H T is tangent to the base circle. Divide the arc T Q 
as at I and J into a convenient number of equal parts, three being 
taken in this illustration. Draw tangents at I and J. Step off the 
distance A T I on 1 1\ thus obtaining 7i, etc. Draw the involute 
curve through A, h, J\, F, thus obtaining the involute pitch surface 
of the cam. The working surface, if a roller is used, is found by 
taking the radius of the roller and drawing a series of arcs with I\, J\, 
etc., as centers and drawing a curve tangent to them as at I2, J2, 
etc. It will be observed that a working curve so drawn will be tan- 
gent to the last construction arc at Q2 on one side of the cam lobe and 



MISCELLANEOUS CAM ACTIONS AND CONSTRUCTIONS 201 

tangent at Qi on the other side, giving the actual tip of the working 
cam surface at Q3. This means thai if the roller follower is to move 
with a velocity characteristic of the involute curve, that its ultimate 
stroke will be less than the desired amount by the distance Q3, Q5. 
This can only be corrected, when a roller follower is used, by disre- 
garding the involute characteristics at the end of the stroke, and by 
arbitrarily changing the true working surface curve from J2 to Q3 
so that the curve will run smoothly from J 2 to Q5. 

402. The involute curve has a fixed initial angle for each 
pressure angle. For example, the initial angle SO A, Fig. 160, 
will always be 39^° for a maximum pressure angle of 30° no matter 
what size of involute is used. This may be readily shown as follows : 
Let y equal the angle S A which is the initial angle from the origin 
of the involute to the point where the pressure angle is to be shown. 
Let a be the assigned pressure angle as represented at H A K. 
Then the angle SOT = y+ (90° - a°). Let x equal O T, the 
radius of the base circle of the involute. Then, x cot a = A T = arc 

S T = 2 7r x '- — — . The value of x cancels, and for a pressure 

OOU 

angle of 30° substituted for a, y is found to be 39^°. Similarly the 
initial angle is found to be 18° for a pressure angle of 40°. 

403. Involute specially adapted for a flat-surface fol- 
lower. The involute curve is naturally adapted for an oscillating 
cam surface where a flat-surface follower is used, and in this case it 
gives a uniform linear velocity to the follower. The natural advan- 
tage of an involute cam for a flat-surface follower, shown in Fig. 162, 
is based on the property of an involute that the tangent X Y to the 
base circle R Q is normal to the involute as at V, and consequently 
that the perpendicular line at T Z is tangent to the involute. There- 
fore, T U may represent the flat surface of a follower collar attached 
to the follower rod S Si. T\ U\ is the follower in its highest posi- 
tion, and T2 U2 in its lowest position, considering that only the part 
A C of the cam surface is used. An involute curve cam as from A to 
C would always be tangent to the flat surface of the follower and the 
line of contact between cam and collar would pass through the center 
of the follower rod, moving up and down between Y2 and Y\. This 
means that there is no pressure angle on the follower rod except that 
due to friction. This last feature of the involute cam gives it, per- 
haps, its greatest practical importance. Where it is desired to give 
the follower a definite velocity and acceleration between its extreme 



202 



CAMS 



points of travel, F2, Y\, the involute cannot be used and the method 
explained in paragraph 76 et seq. must be used. 

404. Oscillating positive drive single-disk cam. This 
cam, illustrated in Fig. 163, might be compared with the yoke cam 
having a swinging follower instead of a reciprocating follower. Its 
method of construction, however, differs from that of the yoke cam. 
In the illustration the oscillating cam L K M receives its motion 
through the link F G, the point F swinging through the arc F\ Fq. 
The follower piece C B D E swings about the fixed center B through 
the angle E\ B E&. The pitch surfaces Cq C\ and Dq D\ are found 
by considering the cam to remain stationary while the follower 
revolves around it in such a way as to retain its relative working 




Fig. 162. 



-Showing that the Involute is Specially Adapted for an OffsET Flat 
Surface Follower 



position at all phases. The detail construction necessary to do this 
is as follows: Through the point B draw the arc B\ Bq to include the 
same angle as the arc F\ Fq. Divide arc B\ Bq into a number of 
equal parts if the shaft A is to turn with uniform angular velocity; 
six parts are used here. With points B\, B2 as centers and distance 
B C as a radius draw short arcs as indicated at G, C2, etc. The 
point C moves in an arc C C" through the same angle as does the 
point E. 

405. If it is desired that the follower move with angular accelera- 
tion and retardation similar to that produced by the crank curve, 
draw the semicircle having C J, Fig. 163, for a radius, divide it into 
the same number of parts as Bu Bq was divided (six in this case) and 
project the division points Ji, J 2, etc., to the arc C" C" which has B 



MISCELLANEOUS CAM ACTIONS AND CONSTRUCTIONS 203 



for its center. Carry these last points around, with A as a center, 
until they meet the corresponding arcs which have beeD already 
drawn with B\ y B2, etc., as centers. Thus, the points C\ . . . Co 
of the pitch surface will be obtained. The points D\ . . . Dq 
of the companion pitch surface are obtained in the same way. The 
radius C J is equal to one-half the chord of the arc C C". Taking 




Fig. 163. — Oscillating Positive Drive Single Disk Cam 

the radius of the roller to be D P the working surfaces A>, M and 
K\ L are obtained. 

406. With the above type of cam, extreme accuracy is necessary 
in manufacture to overcome any binding action of the rollers on the 
cam disk. To overcome this a cam construction has been devised 
in which the two arms B C and B D, Fig. 163, are entirely separated, 
the former being keyed to the shaft B and the latter free to turn on 
shaft B. The two arms are then connected by a spring, as illustrated 
in Fig. 166, which keeps them drawn to each other, and both having 



204 



CAMS 



the desired pressure on the cam surface. To prevent too great a 
pressure of the arms on the cam surface, should too heavy a spring be 
used, a stop pin is cast on each arm and these stops come together 
just as the follower rollers touch the cam surface when newly adjusted. 
407. Cam shaft acting as guide. A special form of construc- 
tion for guiding the cam follower is frequently used as illustrated in 
Figs. 164 and 165. The cam B in Fig. 164 is the simple radial cam 
and is constructed for any given data as explained in paragraphs 49 
et seq. It moves the roller C, which is attached to the forked arm 
R D, back and forth in approximately a radial line the distance A M 
minus A L which is equal to the chord of the arc E D. The arc E D 
measures the swing of the shaft F. The follower rod D R is under 




Fig. 164. 



Fig. 165. 



Fig. 164. — Cam Shaft Guide Takes Place of Crosshead Guide 
Fig. 165. — Positive Drive with Cam Shaft Guide 



definite control all of the time, although its form of construction is 
extremely simple and the number of parts a minimum. The forked 
end R R of the follower rod bears with a snug fit against the two sides 
of the cam shaft, or against adjustable collars attached to the shaft. 
In Fig. 164 the follower shaft F is returned to its initial position by 
means of the spring H. 

408. Positive dkive with cam shaft as guide. A cam giving 
positive motion where the cam shaft is used as a follower guide is 
illustrated in Fig. 165. The cam itself is a face cam and is con- 
structed for any given data as directed in paragraphs 96 et seq. 
The pin C is attached to the follower rod D R and is moved back and 
forth in approximately a radial position by the amount A M minus 
A L, equal to the chord of D E. The forked end of the follower rod, 



MISCELLANEOUS CAM ACTIONS AND CONSTRUCTIONS 205 

bearing against the sides of the cam shaft A, together with the guided 
end D of the rod give it a motion that is under control at all phases, 
and this with a minimum amount of mechanical construction. The 
shaft F is under positive cam control all the time on account of the 
use of the face cam, and no return spring is necessary as in Fig. 104. 

409. Positive drive double disk radial cam w t itii swinging 
follower. A special form of cam and follower construction, where 
positive action is desired, is shown in Fig. 166 where the following 
data are so taken: 

(a) That two follower arms 10 units long shall each swing through 
an angle of 20° with uniform acceleration and retardation while 
two corresponding radial cams turn through 135°, the drive to be 
positive with each roller having a single point of contact. 

(b) That the follower arms shall be returned with positive action 
while the cams turn through 225°. 

(c) That the angle between the two radial follower arms shall 
be 50°. 

410. The follower shaft A, Fig. 166, is first laid down, the angle 
of follower-arm swing of 20° then drawn as at B A C, and finally the 
10 units for follower-arm length laid off at the initial position A B. 
The horizontal centerline E O for the cam shaft is then drawn across 
the arc B C so that the midpoint D is as much above it as the end 
points B and C are below. The radius D of the pitch circle is com- 
puted in the usual way, taking the chord B C for the distance moved 

u +u f ii • + TU n ^ BCX 3.46X360 _ _ xl 
by the follower point. Then D O = = 5.16, thus 

locating the cam center 0. The circle represented by A A\ is then 
drawn and the pitch surface, indicated by the short portion B Bi 
is constructed in exactly the same manner as explained in Problem 8. 
The size of the roller is assumed as shown at B F and the working 
surface of the operating cam F G is drawn. The operating arm A B 
is keyed to the shaft A . 

411. The return arm A H, Fig. 166, is not keyed to the shaft A 
but turns freely on it instead. The motion of this arm should be 
identical with that of the arm A B and, therefore, the swinging arc 
H J of the center of the follower roller is made the same as the arc 
B C, and it is similarly divided. The pitch surface of the return 
cam is represented in part at H Hi and is found in exactly the same 
way as directed in the preceding paragraph, and the working surface 



206 



CAMS 



K L drawn. The spring at M exerts more pull than is required to 
return the follower, and, therefore, it holds the two follower arms 
against the cams with practically uniform pressure, even should 
there be slight inaccuracies in workmanship, or wear in the contact 
surfaces. A lug is attached to each follower arm as shown at P and Q, 




Fig. 166. — Oscillating Positive Drive Double Disk Cam 



and these act as stops in preventing excessive pressure of the rollers 
on the cam surfaces. Cams used in this way have been called duplex 
cams. 

412. Rotary sliding yoke cams giving intermittent har- 
monic motion. A yoke cam driven by sliding contact instead of 
roller contact is shown in Fig. 167. The cam, in this figure, is in the 
form of an equilateral triangle bounded by equal circular arcs having 
a radius equal to the straight sides of the inscribed triangle. The cen- 



MISCELLANEOUS CAM ACTIONS AND CONSTRUCTIONS 207 

ters for the circular arcs are at the apexes of the i riangle. One of the 
apexes of the cam is at the center of the driving shaft. The motion 
given to the follower yoke will be an intermittent one, dwelling at 
the ends of the stroke, and the total travel will be equal to the radius 
of the cam surface. The follower will travel from one end of its 
stroke to the other with a simple harmonic motion the same as with 
crank and connecting rod where the connecting rod is assumed 
to be of infinite length. Or, the motion during the stroke will be the 
same as with the Scotch yoke, or crank and slotted crosshead, where 
the radius of the crank is one-half the radius of the present cam sur- 
face. 

413. A diagram of the motion of the yoke follower in Fig. 167 is 
shown at M N S. With the cam turning as shown by the arrow, the 




Fig. 167. — Sliding Yoke Cam Giving Harmonic Motion 



follower H K will move the distance M N while C on the cam turns 
60° to D, and the cam edge at C will do the driving with a scraping, 
sliding action. While C is turning 60° from D to E, the follower will 
remain at rest; while C is turning 60° from E to F the curved surface 
A C of the cam will be driving the follower the distance P with a 
rubbing, sliding action and increasing velocity; while C is turning 
from F to G the cam edge C will again be driving, the follower moving 
the distance P Q with a scraping, sliding action and decreasing velocity. 
The smooth working surface of the follower yoke is shown from FtoB, 
while the recessed surface as at W may be left rough cast. The 
velocity and acceleration diagrams for the equilateral sliding yoke 
cam here described have the same characteristics as those shown for 
the ordinary crank curve cam, illustrated in Figs. 88 and 89. 

414. Rotary sliding yoke cam giving reciprocating har- 
monic motion. A circle passing through the points A B C, Fig. 167 



208 



CAMS 




Fig. 



168. — Sliding Yoke Cam 
General Case 



would represent the surface of a cam attached to the crank A J, and 
such a cam, instead of the equilateral curved side cam, which is shown, 
would give a simple harmonic motion to the follower yoke without 
finite periods of rest at the ends of the stroke. Such a circular cam 
would be an equivalent of a crank and slotted crosshead where the 
radius of the crank would be equal to the radius of the cam circle. 

415. A ROTARY SLIDING YOKE CAM, GENERAL CASE, With the Cam 

surface entirely surrounding the shaft is shown in Fig. 168. To lay 

out this cam for a definite range of 
motion, say 2 units, draw the indefinite 
circular arc B C with any desired radius 
and the arc E D with the same center 
and with a radius 2 units larger. Then 
with a radius equal to A B plus A E and 
a center anywhere on the arc E D draw 
the arc C D until it intersects E D as at 
D. With Das a center and the same 
radius as before draw the arc E B com- 
pleting the cam. The student should 
be able to determine the angles traveled by the cam while the 
follower is at rest, the angles of motion, the range of motion of the 
follower, and the exact portion of the follower working surface which 
has to resist the wear due to sliding action. 

416. Cam surface on reciprocating follower rod. In some 
special forms of cam construction it is more convenient to place the 
cam curve on the follower than on the driver. Such a case is illus- 
trated in Fig. 169 where the cam curve E F E' is on the sliding fol- 
lower bar K G. The driving crank A F carries a pin at F which slides 
in the cam groove. The mechanism here shown is a modification 
of the Scotch yoke, or " infinite connecting rod." The motion in 
this case is such that the follower remains stationary, while the 
driving shaft turns through the angle C A C. The curve C F C is 
an arc of a circle with A as a center. The follower then picks up 
motion comparatively slowly, the point G being at the points 1,2, 3, 
etc., when the crank pin F is at the points which are correspondingly 
represented in Roman numbers. When the crank pin is at J, G is at 
N and it then moves very rapidly from N to P while the crank pin 
travels from J to Q. Very often, in cam work, the driving shaft A 
has only an oscillating motion through 90° or less. If the curve 
C F C is changed slightly so as not to be an arc of a circle with A 



MISCELLANEOUS CAM ACTIONS AND CONSTRUCTION 2()<) 

as a center, the end G of the follower bar will not come to rest for a 
definite period at the end of the stroke, but it will have a slow, power- 
ful motion which may be made use of in manufacturing processes 
where compression is required. 

417. Problem 42. -Definite motion where cam surface is 
on follower rod. In Fig. 169 a follower rod G K has a cam surface 
formed at the left-hand end from E to E' , and it is driven by a simple 
crank pin represented at F so as to secure a desired or known motion. 
In the illustration let it be desired: 

1st. That the follower rod shall remain at rest at the head end of 
the stroke while the driving crank pin turns 45° (22J^° on each side 
of the centerline A F). 



D 



/niv 



UI 



J\E 
















%-fT 
































vf\_ 


h 




mvif 




P 6 

— 1 — 1 — 


5 


4 

4' 


N 
3 


2 


1 


hri'l 




k' 


1' 


Ljcr 
















1 Ik 






\\e' 

















Fig. 169. — Cam Surface on Reciprocating Follower Rod 



2d. That the follower will be moved to the left a distance G N 
with uniform acceleration while the crank pins turns 67^2- 

3d. That the follower shall move the remainder of the stroke 
from N to P while the crank pin turns 90°. 

4th. That the follower rod shall move in reverse order on the 
return stroke from P to G. 

418. Before starting the solution of this problem it should be 
stated that, one cannot, because of either theoretical or practical 
considerations, or both combined, always secure desired results in 
cams of this type where arbitrary distance and motion assignments 
are given as in this illustration. It is nevertheless advisable to 
complete the solution of the problem, if possible, on the basis of the 
desired data, because one can then make the necessary modifications 



210 CAMS 

with a sure knowledge that the least departure has been made from 
the theoretical or assigned conditions. 

419. The method of solution for the above data is as follows: 
Assume the driving crank length A F and draw the crank-pin circle 
F J M. Lay off the angle F AC equal to 22^°. The circular arc 
F C will then be part of the pitch line of the follower cam head, and 
while the crank pin F is moving through this arc the follower rod will 
not move at all. To secure uniform acceleration of the follower for 
the distance G N, divide G N into 9 equal parts and mark the 1st, 
and 4th division points as indicated at V and 2' in the figure. This 
will be the first step in securing the uniform acceleration called for 
because the distance from Gtol' will be one unit, from 1' to 2' will be 
three units, and from 2' to 3' will be five units. By dividing G N into 
three parts as here described, three construction points will be secured 
on the cam curve. If more construction points are desired, G N 
may be divided in 16 equal parts and the 1st, 4th and 9th inter- 
mediate division points taken, thus obtaining four construction 
points on the part of the pitch surface of the cam from C to E. Like- 
wise, if five construction points are desired, G N would be divided 
into 25 equal parts, and the 1st, 4th, 9th, and 16th division points 
taken. 

420. Since the motion from G to N is to take place while the crank 
pin moves 67^° as called for in the data, and since three construction 
points have been used in this illustration, the 67^° arc from C to J 
is now divided into 3 equal parts as indicated at I and 77 in Fig. 169. 
At I draw a horizontal line and make the distance I-R equal to l'-G; 
at II make the distance IIS equal to 2 f -G; and at J, make the dis- 
tance III-E equal to S'-G. A curve through the points C, R, S 
and E will be the pitch line for the cam surface on the follower rod 
for uniform acceleration from G to N. The p oint 8' coincides with N. 

421. The part of this pitch curve from R to E is shown by a dash 
line and is not practical because of the sharp curvature from S to E, 
which would produce too large a pressure angle and this in turn 
would give a large bending moment on the follower arm and large 
side pressure in the bearing H. This part of the curve should, 
therefore, be modified, and a good plan on which to effect the mod- 
ification is to start by making the pressure angle as large as is prac- 
tically allowable and then to keep the new curve as near to the old as 
possible. A maximum pressure angle that is safe under all ordinary 
circumstances is 30° and, therefore, the first step in the modification 



MISCELLANEOUS CAM ACTIONS AND CONSTRUCTIONS 211 

will be to draw a vertical line through E, the end of the theoretical 
curve, and make an angle of W E V equal to the maximum practical 
pressure angle of 30°. The line V E is then produced until it crosses 
the dash curve, and a smooth curve is next drawn so as to connect the 
straight line and the original curve. This will leave, in this case, 
E T as a straight 30° line, T R as a new assumed part of the pitch 
curve, and R F as the portion of the original curve that remains. If 
the cam is to turn slowly, or if the load on the cam is not large, a 
greater pressure angle could be taken at W E V and then the arbi- 
trary new curve would come closer to the original or theoretical 
curve. 

422. The practical pitch line of the cam is now found to be 
FCRTE. The cam will run smoothly and the variation in the 
motion of the cam from the originally desired motion may be par- 
tially indicated by pointing out that the end G of the follower will be 
at 2, and at 4 instead of 2' and 4' as originally intended. This varia- 
tion may be most completely shown by a velocity diagram which will 
be taken up in a succeeding paragraph. 

423. The pitch curve F T E, it will be noted, has been con- 
structed to give a definite practical action to the follower from G to N. 
Since the curve F E is now determined, and since the crank pin must 
drive through the same cam slot from E to F while it turns through 
the remaining arc J Q, it follows that the motion of the rod from 
N to P cannot be assigned, and that it must be taken as it comes. 
To find out in a general way what this motion will be it is only 
necessary to pursue in reverse order the methods already used; 
i.e., to lay off the distance T-IV at G~4, the distance R-V at G-5, 
etc. By noting the distances N~4, 4~5> etc., which the follower rod 
travels in uniform periods of time, some useful idea of the retardation, 
and consequently of the smoothness of action of the cam may be 
obtained as it approaches the inward end of its stroke. In the 
illustration the follower rod will slow down perceptibly from N to 4, 
and have slightly higher but a fairly uniform velocity from 4 to 5, 
and from 5 to 6. It will retard rapidly from 6 to the end of the 
stroke. 

424. The lower part of the pitch curve from F to E' will be made 
symmetrical with the upper part from F to E in this problem thus 
making the action of the follower on the return stroke the reverse 
of what it is on the forward stroke. If it were desired, the curve 
F E' could be constructed, by the methods described above to give 



212 



CAMS 



the same characteristic motion to the follower on the return stroke 
as it did on the forward stroke. 

425. An exact knowledge of the effect of arbitrarily 
changing the theoretical curve R S E, Fig. 169 to R T E may 



w y 




Fig. 169. — (Duplicate) Cam Surface on Reciprocating Follower Rod 



be readily obtained by a time-velocity diagram construction as 
illustrated in Fig. 170. In the latter figure let the length of the base 
line F Q represent the time necessary for the crank pin to make a half 
revolution from F to Q, Fig. 169. Since the crank pin is assumed 

to travel with uniform velocity, the line 
F Q, Fig. 170, is divided into eight equal 
parts the same as is the semi-circle F Q in 
Fig. 169. The velocity of the follower 
at each of the construction points is 
then found as indicated in the following 
paragraph. 

426. At the point II, for example, in 
Fig. 169, draw the tangential line II-B 
of any desired length. This line will 
represent the velocity of the crank pin 
in feet per second, which may be readily 
computed, for, if the crank A F is 4 inches long and makes 120 revo- 
lutions per minute the point F will be moving with a velocity of 

4 120 

^ X 2 X 3.14 X w = 4.19 feet per second. 




Q6 5 4J21CF 

Fig. 170. — Problem 42, Time- 
Velocity Diagram for Re- 
ciprocating Follower Rod 
Shown in Fig. 169 



MISCELLANEOUS CAM ACTIONS AND CONSTRUCTIONS 213 

Through the point B draw a line B D parallel to the line that is tan- 
gent to the cam pitch curve at T. The line V E, continued, is tan- 
gent to the cam curve at T because it will be remembered that the 
practical curve from R to T was taken so as to be tangent at its upper 
end to the straight line E T. The distance II-D will represent the 
velocity in feet per second with which the follower rod is sliding 
through the bearing at H. This velocity is laid off in the time-velocity 
diagram in Fig. 170 at 2-D. In a similar manner other points on the 
solid-line curve C D Q may be found. This curve shows at a glance 
just how fast the cam follower is moving at every phase of its stroke. 

427. The dash line construction in Fig. 170 shows the follower 
velocities called for in the original data, but abandoned, as explained 
above, because of the large pressure angle involved. The point 
on the dash curve is found by drawing the line B 0, Fig. 169, through 
B parallel to the short straight dash line which is shown tangent to the 
theoretical curve at S. Then II-O would represent the velocity of 
the follower bar at phase 77 if the original data were used. As a 
check on the accuracy of the construction the points C, L, and X, 
Fig. 170, should all be on a straight inclined line, because C X is a 
velocity line and it must show uniformly increasing velocity for the 
follower in order that there may be uniform acceleration as called 
for in the original data. 

428. The difference between the solid and dotted parts of the 
velocity diagram in Fig. 170 shows the effect on the velocity of the 
follower of arbitrarily changing the theoretical cam curve R S E, 
Fig. 1G9, to the more practical cam curve R T E. 

429. Problem 43. Cam surface on swinging follower arm. 
When the cam surface is on the follower and it is desired that the 
follower shall have a swinging motion instead of a rectilinear recip- 
rocating motion as it had in Fig. 1G9, the method of construction 
will vary in detail as illustrated in Fig. 171. The data for Fig. 171 
are, that the driving crank A C with a crank-pin roller at G shall 
swing the follower shaft B through an angle of 30° counterclockwise 
with uniformly increasing and decreasing angular velocity while the 
driving shaft turns through 60° with uniform angular velocity in the 
same direction. 

430. The method of locating points on the curve C F of the fol- 
lower cam pitch surface, Fig. 171, follows: Divide the assigned 30° 
arc, C E, into any number of parts, say six, which are as to each other 
as 1, 3, 5, 5, 3, 1. This will provide for the uniformly increasing and 



214 



CAMS 



decreasing motion to the shaft B. Divide the assigned 60° driver 
arc, C D, into six equal parts. The method of locating the point L, 
which is the second construction point on the cam curve, will be taken 
for explanation purposes. Other points are found in the same way. 
Draw a radial line B 2 through the second construction point, con- 
tinuing it to J which is on an arc which passes through 77 on the arc 
C D . Lay off the arc J K at II-L thus obtaining the point L on 
the cam curve. This form of cam has positive action. When it is 




Fig. 171. 



-Problem 43, Cam Surface on Swinging Follower Arm Having Uniform. 
Angular Acceleration and Retardation 



allowed to reach a dead center position as shown in Fig. 171, auxiliary 
action will be required in starting. 

431. Effect of swinging transmitter arm between ordinary 
radial cam and follower. In Fig. 172 let B C D E F be an ordi- 
nary radial cam with straight sides as at B H rounded off by circular 
arcs with center as at G. Let I J K be the swinging transmitter arm 
with the working surfaces at J and K as arcs of circles with centers 
at L and M respectively. Let N N f be the centerline of the follower 
rod which moves straight up and down. 

432. In order to reach a useful understanding of the action of this 
type of cam construction it will be necessary to learn the rate of 



MISCELLANEOUS CAM ACTIONS AND CONSTRUCTIONS 215 

change of velocities in the follower parts so as to judge the accelera- 
tions and retardations which cause the most trouble at high speeds, 
also to learn the rates of sliding at J and K, and then to balance t hese 
against the pressure angle produced by the same radial cam with an 
ordinary direct roller-end follower. 




Fig. 172. — Swinging Transmitter Arm with Sliding Action 



433. The method of analyzing the cam action in Fig. 172 will be 
pointed out by using six equally spaced construction points during 
the period that the surface B C is in action, the cam turning as shown 
by the arrow. To obtain the positions of the six points for analysis 
one cannot divide the subtended arc B P of the working surface arc 
B C into six equal parts where a swinging follower arm is used, as may 
be recalled from Problems 8 and 11. Instead, it is convenient foi 
analytical construction purposes to revolve the swinging follower 
arm around the cam with uniform angular velocity while the cam 



216 CAMS 

remains stationary. The detail work necessary to accomplish this 
is done first by drawing an arc of a circle through I with A as a center, 
finding where I is on this arc at the beginning and end of action while 
the arm I J slides on B C, and then dividing the arc of swing of I into 
six equal construction parts. 

434. The initial position of I, Fig. 172, is found by laying off the 
distance L J from the point B on the radial line A B, thus obtaining 
the point ; then using as a center and a radius equal to L I draw 
a new arc to intersect the arc through / at 7i. This will be the posi- 
tion of I when the swinging arm is tangent to the cam at B\ in a sim- 
ilar manner 76 will be found to be the position when the arm is tan- 
gent at C. With the arc I\ Iq obtained and divided into six equal 
parts, it is no longer necessary or convenient to consider the center I 
as revolving about A, and it will, therefore, be considered as fixed 
in further work, the next step of which will be to find the six corre- 
sponding positions of the point L. This is readily done by drawing 
an arc through L with I as a center and then taking I L as a radius 
and the point I*, for example, as a center and drawing an arc such as 
one of the short ones shown at O4. Then with a radius equal to L J, 
find by trial, a point on the arc just drawn which will be a center for 
an arc that is tangent to B C of the cam. This center is shown at O4 
and the tangent arc is shown at B4. With A as a center draw an 
arc through the point O4 until it cuts the arc through L already 
drawn, as at L4. In the same manner the six points on the arc 
through L. are found, and the corresponding points of tangency on 
the cam outline B C are obtained as shown from B to C. 

435. The locus of the point of contact may now be found, as 
at R J Re, Fig. 172, as follows: To find, for example, the point #4, 
draw two intersecting arcs, one having L J for a radius and L\ for a 
center and the other having A B± for a radius and A for a center. 
Similarly other points on R J R§ are found. 

436. The angular velocity curve for the swinging follower 
arm may now be readily found and its acceleration and retardation 
judged. Let S T represent the linear velocity of a point S at radius 
A S on the cam. Then a point at B± on the working surface of the 
cam has a linear velocity of S' T' and this value is laid off at R± T2 
where the point B± is in action. The component of R± T2 that pro- 
duces rotation in the swinging follower arm is R± T3, perpendicular 
to I #4, and this reduced to a radius of I $4, equal to A S, for purpose 
of comparison with the cam rotation, is $4 T4. This value is laid 



MISCELLANEOUS CAM ACTIONS AND CONSTRUCTIONS 217 

off on the 4th ordinate in the velocity diagram in Fig. 173, as at 
#4 7 7 4. In a similar manner other values are obtained in Fig. 173 
and the curve B Q C drawn. This curve shows the rate of change 
of angular velocity in the transmitting follower arm while the straight 
horizontal line D E shews the uniform angular velocity of the driving 
cam. The length B C of the base line of the velocity diagram may 
be taken any length and then divided into six equal parts to locate 
the various ordinates of the velocity diagram. The length D B in 
Fig. 173 equals S T in Fig. 172. 

437. The amount of sliding of the cam may readily be found 
for example, by first breaking up the velocity R± T2 of the point R± 
on the cam in Fig. 172 into its normal and tangential components — 
the former being shown at R± T5 and the latter at R± Ta — and, second, 
by breaking up the velocity R± T3 of the corresponding point on the 
swinging arm into the components Ra T5 and R± T7. The difference 
Tq T7, in the longitudinal components will be the rate of sliding at 
that phase and this difference is laid off at S T in Fig. 174. Sim- 
ilarly other points on the curve D T F are found. The rate of sliding 
when the circular surface of the cam B F E is in rction, providing 
there is no stop rest for the follower arm, is B D, equal to S T in Fig. 
172; and when the surface C D is in action it is C 1\ Fig. 174. 

438. A VELOCITY CURVE FOR THE FOLLOWER ROD N N' , Fig. 172, 

will give some indication of its acceleration and retardation and the 
relative strength of spring required to operate it in comparison with 
the results secured by an ordinary roller-end follower. The first 
step in this construction consists in finding the six positions of the 
center M of the upper curved surface K' K" of the swinging arm. 
This is readily done because the points M, L and / are fixed relatively 
to each other, and, therefore, the point M4, for example, is found by 
taking / M as a radius, / as a center, and drawing an arc at Mo M&. 
Then with L M as a radius and L4 as a center draw another short 
arc intersecting the first, as at M4. With M K as a radius and M± as 
a center draw the arc passing through ivV The point K4 is on a ver- 
tical line through A/4. The horizontal line tangent to the arc at 
K4 will have the position of the bottom of the follower rod at phase 4- 
In a similar way other points on the curve K$ K$ which is the locus 
of the point of tangency, is obtained. The distances between the 
horizontal lines drawn through the points Kq, K\, etc., will show the 
amount of vertical travel of the follower during each of the six equal 
time periods. 



218 



CAMS 



439. The velocity diagram for the vertical follower rod is quickly 
obtained by first laying off the same angular velocity for the swinging 
arm at K4, Fig. 172, as was found at R± and finding the vertical com- 
ponent of the velocity of the point K±. This is done in detail by tak- 
ing the unit radius I $4 together with the linear velocity $4 T± at 
this unit radius, both of which have already been found, and trans- 




Fig. 172. — (Duplicate) Swinging Transmitter Arm with Sliding Action 



ferring the distance $4 T4 to £5 TV This will represent the linear 
velocity of the point £5 on the radial line I K±. The resultant linear 
velocity of the contact point K± on the swinging arm is found, as 
shown, to be equal to K± Tg. The vertical component K4 T\o of 
this resultant velocity for the arm gives the actual upward velocity 
of the rod TV N'. This value is laid off at S Tio in Fig. 175 and is an 
ordinate on the velocity curve B Q C. 



MISCELLANEOUS CAM ACTIONS AND CONSTRUCTIONS 219 



440. The corresponding ordinate S U3, Fig. 
175, for the velocity curve of the follower rod, 
if it had an ordinary roller end with a roller 
radius equal to B B' of Fig. 172, may be found 
1st, by drawing the pitch surface line B' C of 
the cam; 2d, by dividing the arc B P into six 
equal parts; 3d, by drawing a radial line through 
the fourth point P4 to B\\ 4th, by revolving 
B\ to N4 and obtaining the full linear velocity 
N4 T± and laying it off at B\ U; 5th, by find- 
ing the radial velocity B\ U2 by drawing the 
line U U2 perpendicular to the normal B\ JJ\. 
The length B\ U2 will then represent the 
velocity of the follower bar if it had a roller 
end and this length is laid off at S U3 in Fig. 
175. Similarly other ordinates of the curve 
B UsC are found. 

441. Comparing the velocities of the 
follower rod N N', Fig. 172, when a trans- 
mitting swinging arm is used and when an 
ordinary roller end is used, it will be seen that 
the follower rod attains a higher velocity in the 
former case as shown by the greater height of 
the curve B Q C over the curve B U3 C. Also 
the acceleration of the follower rod N N' on the 
upstroke will be greater with the swinging arm 
as is indicated by the greater steepness of the 
curve from B to Q over that of the curve from 
B to U 3 . 

442. The sliding action of the surface 
K f K", Fig. 172, of the swinging follower arm 
on the bottom of the rod A7 N' has a max- 
imum value of about one- fifth of that of the 
cam surface B C on the lower face of the 

Fig. 173. — Angular Velocity Diagram for Cam and 
Swinging Arm 
Fig. 174. — Sliding Velocity Diagram of Cam on Swing- 
ing Arm and of Arm on Follower Rod 
Fig. 175. — Linear Velocity of Follower Rod, with Trans- 
mitting Arm and with Ordinary Roller Follower 
Fig. 176. — Pressure Angle Diagram, with Ordinary 
Roller Follower 



£ 900- 
^760- 

Seoo- 

eo 
*450 

O 

F 300- 



s 



k 



B S 4 

Fig. 173. 



600n 




Fig. 175. 




Fig. 176. 



220 CAMS 

swinging arm. This is readily determined by making use of work 
already done, as, for example, by simply measuring the line 
T 9 Tio, in Fig. 172, which is the horizontal or sliding component 
of the resultant velocity K± Tg when the point of driving contact is at 
IQ. The distance T 9 T 10 is laid off at S T n in Fig. 174. Other 
points of the curve B Tn C are found in the same way. The ordinates 
of this curve added to those of the curve D T F would give a measure 
to the total sliding action at any instant when a swinging transmitting 
arm is used. 

443. If an oedinary roller follower instead of a swinging 
transmitting arm were used the pressure angle which would exist, 
with a cam of the size used in Fig. 172 and with a radius of roller equal 
to B B', may also be readily determined from work already done. 
For example, when the center B\ of the roller is in action the roller 
will be pressing against the cam in the direction of the normal B\ L\ 
relatively to the cam and the follower rod will be moving in the direc- 
tion of the radial line B\ U2 relatively to the cam. Therefore the 
pressure angle at phase 4- would be a, which is equal to 29°, and this 
value is laid off on the fourth ordinate as at S V in Fig. 176, thus ob- 
taining a point on the pressure angle curve which, it will be noted, 
has a maximum of about 31° — a very easy angle for general use. 

444. If it is desired to know the actual rubbing velocities 
in feet per minute of the cam on the swinging arm, and of the arm 
on the follower rod; and the linear velocity of the follower rod N N f , 
Fig. 172, it may quickly be obtained from the velocity diagrams now 
drawn, for any given problem. For example, let it be assumed in 
this problem that the short radius A B of the cam in Fig. 172 is % 
inch and that the cam is making 900 revolutions per minute. 

445. For the data just assumed the point B on the cam will 

1 - + t . 75 X 2 X 3.14 X 900 
be moving with a velocity 01 r~ = 006 teet per 

minute. This then would be the velocity represented by the line 
S T in Fig. 172. Since all the velocity lines shown in the drawings 
have been found and laid down without any change in the scale of 
the drawing, it is only necessary to compute the distance on S T that 
represents 100 feet per minute, and to make that distance the unit 
for the velocity scale for measuring the curves in Figs. 174 and 175. 
If A S measures % inch, S T will be found to measure .98 inch to 
the same scale. Then .98 inch represents 353 feet per minute, or, in 
other words, .28 inch represents 100 feet per minute. In Figs. 174 



MISCELLANEOUS CAM ACTIONS AND CONSTRUCTIONS 221 

and 175 the distance BA is .28 inch to the same scale on which 
A B was measured in Fig. 172, and this distance becomes the unit 
measurement for 100 feet per minute in the velocity diagrams. 

446. By drawing the scales as above described it will be noted, 
in Fig. 174, that the maximum rubbing velocity of the cam on the 
lower face of the swinging arm is about 560 feet per minute and that 
the maximum rubbing velocity of the upper face of the swinging arm 
on the bottom of the follower rod is about 110 feet per minute. These 
considerations would affect the design in so far as lubrication and 
wear arc concerned. 

447. The maximum velocity of the follower rod N N' 
in Fig. 172, may also be read off directly in Fig. 175, after the scale 
has been laid down as above described. This maximum velocity, it 
will be noted, is about 460 feet per minute. Had an ordinary roller 
follower been used on the end of the follower rod, the maximum 
velocity of the rod would have been appreciably less, or about 380 
feet per minute. This consideration has an important bearing on 
strength of the moving parts in the general design of cam work. Its 
comparative effect, as for example in the strength of spring required 
to return the follower parts, may be definitely obtained by con- 
structing an acceleration and retardation diagram from the velocity 
curves shown in Fig. 175, as explained in detail in paragraph 268, 
et seq. 

448. Boundary of surface subject to wear. In a cam design 
where there is a sliding follower as in Fig. 172 it will be of advantage 
to know not only the rubbing velocities as found above, but also the 
limits of the surfaces on which the rubbing takes place and the posi- 
tions on the surfaces where the rubbing velocities arc highest and the 
pressures due to acceleration are greatest. With respect to the cam in 
this problem, the conditions are ideal because the accelerations of 
the follower parts are greatest when the rubbing velocities are least. 
This combination occurs on the portion of the cam surface between 
B\ and Bi as may be pointed out as follows: (a) In Fig. 175 the 
velocity curve B Q is steepest between the phases 1 and 2 and con- 
sequently the acceleration of the follower rod N N' is greatest; (b), 
In Fig. 173 where the angular acceleration of the swinging arm is 
greatest also between 1 and 2\ (c), and in Fig. 174 where the sliding 
velocity is lowest between 1 and 2. The conditions for the swinging 
arm I are not so good. In the first place the total wear on the lower 
surface of the arm on the upstroke takes place between J' and J-$ as 



222 cams 

found by drawing the dashline arcs through the extremities R, R3 
and Rq of the path of action taking / as a center in each case; sec- 
ondly, the portion of the surface from Jq to J3 is rubbed over twice on 
the upstroke, or, in other words it receives twice as much wear as the 
part from J' to Jq; thirdly, the rubbing velocities are highest while 
the doubly worn surface from J3 to Jq is in action as indicated by the 
higher part of the curve from Q to F in Fig. 174; fourthly, the part 
of the swinging arm surface just to the right of Jq is also under the 
most intense pressure, due to acceleration, as well as being subjected 
to double wear and high velocity, as may be noted by the fact that Jq 
lies between the phases Ji and J 2 and that between these phases the 
accelerations are greatest, as indicated by the steepness of the curves 
between the ordinates 1 and 2 in Figs. 173 and 175. The points J\ 
and J 2 are not shown in Fig. 172, but they may be readily found by 
drawing arcs through R\ and R 2 with I as a center. The point R2 is 
on the path of action just above the point B2. 

449. Cam action different on up-and-down strokes. All of 
the velocity and sliding curves obtained as above for the cam with 
a transmitting swinging arm, it will be noted, are for the action that 
takes place while the follower rod N N', Fig. 172, is on its upstroke, 
or, in other words, while the part of the cam surface from B to C is in 
action. While the follower is on its downstroke the surface of the 
cam from D to E is in action and the velocity and the sliding curves 
will be different, and should be obtained by similar methods where 
full information for specific practical application is desired. It 
may easily happen, according to the forms of the acting faces of 
the swinging arm, that the velocities and the accelerations and retard- 
ations may be quite different on the two strokes. Hence the informa- 
tion regarding both strokes should be known in order to properly 
judge the friction and wearing characteristics, and also to judge the 
strength of parts to be used. 

450. The disadvantage of the side pressure that accompanies the 
ordinary roller-end follower, and the disadvantage of the high rubbing 
velocity that accompanies the swinging transmitter arm which is 
illustrated at / J K in Fig. 172, may be overcome by using a roller on 
the swinging arm to act against the surface B C of the cam, and a 
roller on the end of the follower arm to act on the transmitter head 
at K' K" . The side pressure produced by the slope of the cam is thus 
taken up by a tensional strain in the swinging arm instead of a side 
strain in the follower rod N N', and a smoother and easier cam action 



MISCELLANEOUS CAM ACTIONS AND CONSTRUCTIONS 223 

should result although there will be an increased number of parts in 
the cam mechanism. 

451. Problem 44. Small cams with small pressure angles 

SECURED BY USING VARIABLE DRIVE. By giving the cam sliafl a 

variable angular velocity very quick follower action may be secured 
with a relatively small cam without appreciably increasing the pres- 
sure angle. To illustrate, the same data will be taken as were used in 
Problem 3 except that the follower is to move up the given 3 units in 
45° instead of 90°. The complete statement of the present problem 
is as follows: Required a single step radial cam to move a follower 
3 units in 45° turn of the main shaft with uniform acceleration and 
retardation; to similarly return it in the next 45°, and to allow it to 
rest for the remainder of the cycle. 

452. Let N, Fig. 177, be the center of the uniformly rotating 
main shaft of the machine to which the cam is to be applied. Assume 
any length for the driving arm N P and draw the two 45° angles 
P N T and T N Q. Draw the circle whose radius is N P and divide 
each of the arcs P T and T Q into six equal parts. Connect the 
points Q and P, thus obtaining the point on N T which will be 
the center of the auxiliary or cam shaft. Attach a slotted arm H 
to the cam shaft, making the shorter working radius of the arm J 
equal to T, and the longer working radius H equal to N plus 
N P. Assume the diameter of the driving pin at P which works in 
the slotted arm, and make the length of the slot a little greater 
than J H to allow for clearance. 

453. Variable drive by the Whitworth motion. From the 
preceding paragraph it may now be seen that the arm H, Fig. 177 
and the cam shaft to which it is keyed will turn through 90° while the 
main machine shaft turns through 45°. The mechanism thus far 
described for producing this result is equivalent to the Whitworth 
slow-advance and quick-return mechanism, but any other type of 
slow-advance and quick-return mechanism that gives complete 
rotary motion could be used instead. 

454. To construct the cam, compute the size of the pitch circle 
in the same manner as in an elementary problem, but using the 90° 
that the cam w r ill turn during the outward motion of the follower 
instead of the assigned motion of 45° that the main shaft will turn. 
Thus the diameter of the pitch circle will be found to be, 

3 X 3.46 X 360 



3.14 X 90 



13.2. 



224 



CAMS 



Lay this value off at D S, Fig. 177, and draw the pitch circle with 
as a center. Lay off the assigned motion of 3 units of the follower 
symmetrically about D as at A V. Assuming 6 construction points 
for finding the cam pitch curve, divide A D into nine equal parts and 
take the 1st, 4th and 9th division points; do the same with V D. 
Divide the arc Q T into six equal parts and draw radial lines through 
each division point, as indicated at E and K. Carry the division 




Fig. 177. Problem 44, Showing that Very Small Cams and Small Pressure Angles 
May be Obtained by Using Variable Velocity Drive 



points on Fi around to their corresponding radial lines by means 
of circular arcs, as indicated at A K\ . Then the curve through the 
points A, Ki, etc., will be on the pitch surface of the desired cam. 

455. The present cam does the same work in half the time of 
the cam that is shown in Fig. 32, and both have the same overall 
dimensions. The cams are of different shape, however. The cam 
shaft will have widely varying angular velocity, ranging between 



MLSCELLANEOUS CAM ACTIONS AND CONSTRUCTIONS 225 



values which vary from — - to — -. At the phase of the mechan- 
ic J . U 11 

ism shown by the object lines in Fig. 177 the driving shaft N and 
the cam shaft have the same angular velocity, and this is true 
for this phase no matter what length of driving arm is taken at the 
start. The cam will have its greatest angular velocity when N P 
is in the position N T, but at this phase the pressure angle will be 
zero, and it will be comparatively small while the cam is approach- 
ing and receding from this phase. Had a cam been constructed in 




Swash-plate Cam 



the regular way, that is without variable drive of the cam shaft, to 
give 3 units motion in 45° under the condition of this problem it 
would have required a cam with a pitch circle diameter of 

3 X 3.46 X 360 



3.14 X 45 



26.4 



units instead of 13.2 as here used. 

456. Swash-plate cams differ in structural details from any 
thus far considered but they are, in effect, end or cylindrical cams. 
If in Fig. 178 a basic cylinder C Y is intersected by an inclined 
plane P L it will cut a flat surface from the cylinder and the form 



226 cams 

of this surface, when viewed perpendicularly, will appear as an 
ellipse in which the major and minor axes will be PL and N L, 
respectively. The flat surface thus formed is termed a swash plate. 
It is shown in the top view by the elliptical curve P2 L2 and in the 
end view by the circle Pi L\. As the swash plate turns on its axis 
X X\, which is the axis of the original cylinder it gives a reciprocating 
motion to a follower rod F E. The range of the follower motion will 
be greater or less according to the true radial distance A\ D\ of the 
follower from the axis of the cam, and in the present illustration the 
range of follower motion is R S. If a sharp F-edge were used on 
the follower E F, the contact would be at A, instead of at T as it is 
with the roller, and the motion of the follower would be harmonic, 
giving velocity and acceleration curves similar to those shown in 
Figs. 88 and 89 respectively. The smaller the follower roller F T, 
Fig. 178, the truer and smoother will be the running of the swash- 
plate cam. 

457. Rotary cam giving intermittent rotary motion. A 
cam of unusual form is shown at A B in Fig. 179. It is designed to 
change a uniform rotary motion in the shaft P P to an intermittent 
rotary motion in the shaft C by operating on the roller pins D, E, F, G. 
Specifically, it is desired that the shaft C shall make a 34 turn while 
the shaft P P makes a Y2 turn, then that the shaft C shall remain 
stationary while P P makes Yi turn, and finally, that shaft C shall 
be under positive control all the time. Such a cam would be auto- 
matically formed on a previously prepared blank by using a rotary 
cutter of the same size as the rollers, D, E, etc., and which travels in 
the same path as the rollers while the cam blank is turned by inde- 
pendent means. For the purpose of laying out the blank and of 
representing the form of the cam surface in a drawing, the roller is 
taken in several intermediate positions, one of which is shown in fine 
lines at H, and constructions made as follows. The method here 
given is reduced to simplest terms and is approximate. It is suf- 
ficient, however, for the cam surface will be true, because of its auto- 
matic manufacture, even if the delineation is not exactly so. 

458. Make an end view of the roller as shown at Hi, Fig. 179. 
The circle through Hi represents the circle half way down the roller 
and the short vertical line tangent to it locates the point of tangency 
for the cam surface and roller assuming that the cam surface has a 
45° slant when the roller has turned 45° from E to H. Projecting 
Hi first to H and then down to H2 on the centerline, a point will 



MISCELLANEOUS CAM ACTIONS AND CONSTRUCTIONS 227 

be found on the centerline of the cam surface, it being noted thai 
the cam turns through 90° while the follower turns 45°. A straight 
line on the surface of the roller through H would represent approxi- 
mately the line of contact between cam surface and roller and would 
be projected down to give K and N if the slant of the cam surfaces 
edges were the same. The slant for the edge J K L of larger radius 
is a little less than that of M N 0, being on a larger average radius, 
and, making a corresponding allowance, H2 K is taken a little less 




l 'o 

Fig. 179. — Special Cam Type Giving Intermittent Rotary Motion 



than H 2 N. The points L and will be directly under the roller in 
position F, and the distances from the centerline P P to these points 
will be the same as the distances from the centerline to the corre- 
sponding points on the bottom of the roller. It will be noted in this 
construction that the width of the cam surface is less than the length 
of the roller. 

459. The eccentric may be considered as a special type of cam. 
It is widely used in engine and other work where it is desired to 
secure a simple reciprocating motion from a rotary motion. Where 



228 



CAMS 



the initial motion may be taken from the end of a rotating shaft, a 
crank is the simpler device to use, but where the motion must be 
taken from an intermediate point on the shaft an eccentric is neces- 




Fig. 180. — Practical Example of Cam Shaft Carrying Eleven Cams 

sary. The eccentric gives a characteristic motion to the follower 
the same as a driving crank would give to the crosshead in an ordinary 
crank and connecting-rod mechanism, the equivalent crank length 
being equal to the distance from the center of the shaft to the center 




Separate End Views of Cams Shown in Fig. 180, to Reduced Scale 



of the eccentric circle as shown at R S in Fig. 181. The eccentric 
cannot be used where specific intermediate velocities are desired for 
the follower. The use of the eccentric as a cam in automatic machin- 
ery is illustrated in Fig. 180 which represents the main cam shaft of a 
machine devised for special manufacturing purposes. Eleven cams, 



MISCELLANEOUS CAM ACTIONS AND CONSTRUCT IONS 229 

compactly arranged, are shown on this shaft, four of them being 
eccentrics, namely Nos. e, g, h, and k. All eleven cams are shown in 
end view in Fig. 181 with the exception of i, which is shown to en- 
larged scale in Fig. 179. 

460. An example of a time-chart diagram for all of the cams 
illustrated in Figs. 180 and 181 is given in Fig. 182. Time-chart 



1 

2 


Bunch Plunger 
Cam 

Ba«k Fold 
Cam 
















"111 


1 Ml 1 1 ±d= 


i-.-i-: _______ 




















. 


. ■ ■ ■ " '" — — ] 


3 


Placer Cam 




































^ 


4 


Ink Roll 
Cam 












| ' 




























1 1 


















































fr— y*"fl 






^t4J-j 
































5 


Seal Fold 




- 




1_ 




L ! 1 1 i 1 1 ~ 












6 


Seal Fold 
Cam 


— — 














































1 ; 


























— 








~ 


= 


— 










— 






































































7 


End Fold 
Eccentric 


























































- — — 












— I'll 












































































































8 


Creaser 
Eccentric 






- - 


r*^g- 






































































— — ; — 


9 


Main Cam 


























































































=p 
































































































1C 


Printing 
Cam 














































































































































11 


Rock Shaft 
Eccentric 




















































-L 



Fig. 182. — Practical Example of Time Chart Diagram for Eleven Cams in Oni 

Automatic Machine 



diagrams are treated in a general way in paragraph 19, and in detail 
with reference to a specific example in paragraphs 143 to 147. 
The form of diagram here shown is specially to be commended in 
that the individual diagram boxes for each cam are separated from 
each other by a small space so that it is impossible for the heavy 
base lines to touch or cross each other under any circumstances. 



INDEX 



A PAGE 

Acceleration diagrams for different 

base curves 89 

Acceleration diagrams. Method 

of determining 138 

Accelerations produced by differ- 
ent base curves 142 

Accuracy in cam construction. . . . 146 

Adjustable cam defined 11 

Adjustable cylindrical cam plates. . 193 

All-logarithmic base curve 89 

All-logarithmic cam problem 94 

Angular velocity curve for swing- 
ing follower 216 

B 

Balancing of cams 148 

Barrel cam defined 7 

Base curve defined 14 

Base curves in common use 14 

Base curves. Comparison of . . . . 88 

Base curves. Complete list of . . . 88 
Base curves. Construction of 

common 20 

Base line defined 14 

Box cam defined 8 

C 
Cam action different in up-and- 
down strokes 160, 222 

Cam chart applied 29 

Cam chart defined : 12 

Cam chart diagram defined 12 

Cam considered as bent chart .... 34 

Cam defined 1 

Cam factor chart for all base 

curves 151 

Cam factor chart for common base 

curves 19 

Cam factors for all base curves .... 150 
Cam factors for common base 

curves 18 

231 



PAGE 

Cam factors. Method of deter- 
mining 152 

Cam mechanism for drawing 

ellipse 79 

Cam mechanism for reproducing 

designs 80 

Cam shaft acting as guide 204 

Cam size. Effect on pressure 

angle 33 

Cam surface on follower 208, 213 

Cam with flat-surface follower. ... 45 

Cam with sliding follower 57 

Cams classified 1 

Cams for high-speed work 148 

Cams for low-starting velocities 

129, 132 
Cams for swinging follower arms 

50, 52, 57 

Carrier cam defined 11 

Characteristics of base curves .... 88 

Circles. Subdivision of 86 

Circular base curve. Case 1 119 

Circular cam problem. Case II... 129 

Clamp cam defined 11 

Comparison of base curves 88 

Comparison of parabola and crank 

curves Ill 

Comparison of velocities and forces 

of different base curves 141 

Conical cams defined 2 

Conical follower pin for cylindrical 

cam 190 

Construction of common base 

curves 20 

Crank curve as projection of helix. . 108 
Crank curve characteristics. . .108, 111 

Crank curve construction 21 

Cube base curve 125 

Cube curve cam problem. Case 1 . 127 
Cube curve cam problem. Case 

II 133 



232 



INDEX 



PAGE 

Cube curve cam specially adapted 

for follower returned by spring 144 

Curved follower toe 162 

Cylindrical cam defined 1,7 

Cylindrical cam problem 68, 70 

Cylindrical cams. Drawing of 

grooves in 186 

Cylindrical cams. True pressure 

angle in 186 

D 

Derived curve for pure rolling 

action 174 

Diagram. Cam chart 12 

Diagram. Timing 13 

Disk cam defined 1 

Dog cam defined 11 

Double-acting cam defined 9 

Double-disk positive drive cam 

for swinging arms 205 

Double-disk yoke cam problem .... 65 

Double-end cam defined 7 

Double-mounted cam defined. ... 11 

Double-screw cams 194 

Double-step radial cam 39 

Drum cam defined 7 

E 

Eccentric as a cam 227 

Ellipse. Cam mechanism for 

drawing of 79 

Ellipse. Construction of 178 

Elliptical arcs for pure rolling 

action 177 

Elliptical base curve character- 
istics 123 

Elliptical curve construction 23 

Empirical cam design 25 

End cam defined 7 

F 

Face cam defined 3 

Face cam problem 55 

Factors. Methods of determining 

cam 152 

Factors. Table of cam 18, 150 



PAGE 

Flat-surface follower 45, 49, 59 

Follower carrying cam surface . 208, 213 
Follower returned by springs .... 142 
Follower rollers for cylindrical 

cams 188 

Follower roller. Size of 35 

Follower velocity in ft. per sec. 

165, 220 

Follower with curved toe 162 

Forces produced by different base 

curves 141 

Formula for cam size 17 

Frog cam defined 2 

G 

Gradual starting of follower shaft. . 177 
Graphical methods. Degree of 

precision in 141 

Gravity curve 110 

Groove cam defined 7 

H 

Handwriting. Cam mechanism 

for reproducing 79 

Harmonic curve 108 

Harmonic motion 108, 206 

Heart cam defined 3 

Helix as pro j ection of crank curv e . . 1 08 

High speed in cam work 148 

Hyperbola for pure rolling action . 184 
Hyperboloidal follower pin for 

cylindrical cam 190 

I 

Infinite connecting rod 108 

Interference of cams 75 

Intermediate transmitter arm .... 214 

Intermittent harmonic motion.. . . 206 

Intermittent rotary motion 226 

Internal cam defined 8 

Involute cam problem 199 

Involute curve defined 197 

Involute curve. Construction of. 192 

K 

Keyways. Location of 78 



INDEX 



233 



L PAGE 

Length of follower surface 

58, 62, 159, 164 
Limited use of flat-surface follow- 
ers 49, 59 

Limited use of single-disk yoke 

cams 64 

Limiting size of follower roller 35 

Locus of point of contact between 

cam and follower 

58, 62, 159, 164, 216 
Logarithmic-combination cam 

problem 101 

Logarithmic curve. Construction 

of 101, 172 

Logarithmic curve for pure rolling 

action 169, 171 

Logarithmic curve. Properties of 171 
Logarithmic spiral. Construction 

of 95,98 

M 

Multiple-mounted cam defined .. . 11 

Mushroom cam defined 3 

Mushroom cam problem 45 

N 

Names of cams tabulated 12 

Noise from cams 147 

O 

Offset cam defined 8 

Offset cam problem 42 

Omission of cam chart 31 

Oscillating cam defined 11 

Oscillating single-disk positive- 
drive cam 202 

P 

Parabola cam characteristics 110 

Parabola construction 22, 182 

Parabola for pure rolling action. . . 182 

Parabolic easing-off arcs 103 

Parabolic curve. Property of . . . . 182 

Perfect cam action 110 

Periphery cam defined 2 

Pins for cylindrical cams 188 

Pitch circle defined 16 



PAGE 

Pitch line defined 15 

Pitch point defined 16 

Pitch surface defined 16 

Plate cam defined 3 

Plates for cylindrical cams 193 

Positive-drive cam defined 8 

Positive-drive double-disk cam for 

swinging arms 205 

Positive-drive single-disk cam for 

swinging arms 202 

Precision of graphical methods ... 141 
Pressure angle characteristics of 

involute curve 198, 201 

Pressure angle defined 16 

Pressure angle factors ... 18, 149, 150 
Pressure angle relation to cam size . 31 
Pure rolling in cam work 168-185 

R 

Radial cam defined 1 

Radius of curvature of non-circular 

arcs 38 

Rate of sliding of cam on surface of 

follower 164, 166 

Regulation of noise in cam design 147 
Relative strengths of springs re- 
quired for different cams .... 143 

Roller. Limiting size of 35 

Rollers for cylindrical cams 188 

Rolling action 168-185 

Rolling cam defined 5 

Rotary cam giving intermittent 

rotary motion 226 

Rotary sliding-disk yoke cam . .205, 206 

S 

Scotch yoke 207 

Screw cams 193 

Shaft guide for cam followers .... 204 

Side cam defined 1 

Sine curve 108 

Single-acting cam defined 9 

Single-disk positive drive cam for 

swinging arms 202 

Single-disk yoke cam problem .... 63 

Single-step cam problem 28, 31 

Sinusoid 108 



234 



INDEX 



PAGE 

Sliding contact follower 57 

Sliding friction eliminated 168 

Sliding of cam on follower surface, 

164, 166, 219 
Slow-advance and quick-return by 

cylindrical cams 195 

Small cams with small pressure 
angles secured by variable 

speed drive 223 

Spherical cam defined 2 

Springs. Use of, for returning 

cam followers 142 

Starting velocities of cam followers 

129, 132 

Step cam defined 9 

Straight-line base curve construc- 
tion 20 

Straight-line base curve problem . 106 
Straight-line combination base 

curve construction 20, 107 

Straight-sliding plate cams 196 

Strap cam defined 11 

Subdivision of circles 86 

Sub tangent of logarithmic curve... 102 

Swash plate cam 225 

Swinging follower arm 50 

Swinging transmitter arm 214 

T 

Table of cam factors for all base 

curves 150 

Tangential base curve 113 

Tangential cam problem. Case I. 113 
Tangential cam problem. Case 

II 135 

Technical cam design 27 

Time-acceleration diagrams 139 

Time chart applied 76 

Time chart defined 13 



Time-chart diagram for eleven 

cams 229 

Time-distance diagrams 138 

Time-velocity diagrams 138 

Timing of cams. Problem 75 

Toe-and-wiper cam defined 7 

Toe-and-wiper cam problem 61 

Toe-and wiper cam with variable 

angular velocity 157 

Transmitter arm between cam and 

follower 214 



Variable angular velocity in cam 

shaft 157 

Variable speed for small cams .... 223 
Varied forms of fundamental base 

curves Ill, 149 

Velocities produced by different 

base curves 141 

Velocity diagrams for different 

base curves 89 

Velocity diagrams. Method of 

determining 138 

Velocity of follower in feet per 

second 165, 220 

W 
Wear. Distribution of, on follower 

surface 58, 62, 159, 164, 221 

Whitworth motion 223 

Wiper cam defined 5 

Working surface defined 16 



Yoke cam defined 6 

Yoke cam with rotary sliding disk 206 
Yoke cam problem. Double-disk . 65 
Yoke cam problem. Single-disk.. 63 




Wiley Special Subject Catalogues 

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are arranged in groups — each catalogue having a key 
symbol. (See special Subject List Below). To 
obtain any of these catalogues, send a postal using 
the key symbols of the Catalogues desired. 



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Canning and Preserving. 

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Production; Paint; Printing; Sugar Manufacture; Textile. 

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and Pharmaceutical; Su^ar. 

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Supply. 



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and Miscellaneous Apparatus. 



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